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topchildadoptionguide · 11 months

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How to solve for y

Solving for 'y' is a fundamental skill in mathematics that is often encountered in algebraic equations and functions. Whether you're a student learning the basics or an adult refreshing your math skills, understanding how to solve for 'y' opens the door to a deeper comprehension of equations and their solutions. In this article, we will provide you with a step-by-step guide on how to solve for 'y' in various scenarios, equipping you with the tools to tackle equations with confidence.

Understand the Equation Type: Before diving into solving for 'y', it's crucial to identify the type of equation you're working with. Equations can be linear, quadratic, exponential, logarithmic, or trigonometric, among others. Each equation type requires specific techniques and strategies for solving. Familiarize yourself with the characteristics of the equation at hand to determine the appropriate approach.

Isolate 'y' on One Side: The goal is to isolate the variable 'y' on one side of the equation. To achieve this, use algebraic operations such as addition, subtraction, multiplication, and division to move terms and constants from one side of the equation to the other. Perform the same operation on both sides to maintain equality. Continue simplifying the equation until 'y' is isolated.

Solve Linear Equations: In linear equations, 'y' typically appears as a coefficient multiplied by 'x' or other variables. To solve for 'y', isolate the term containing 'y' on one side of the equation by applying inverse operations. For example, if 'y' is multiplied by a coefficient, divide both sides of the equation by that coefficient. Simplify further until 'y' is isolated.

Solve Quadratic Equations: Quadratic equations involve 'y' raised to the power of 2. To solve for 'y', rearrange the equation into the standard quadratic form (ax^2 + bx + c = 0). Utilize factoring, completing the square, or the quadratic formula to find the values of 'y'. Be mindful of possible solutions, including real and complex numbers.

Solve Exponential and Logarithmic Equations: Exponential and logarithmic equations involve 'y' as the exponent or argument. To solve for 'y', use logarithmic properties to transform the equation into a manageable form. Apply inverse operations, such as exponentiation or taking the logarithm, to isolate 'y' and find its value.

Solve Trigonometric Equations: Trigonometric equations involve 'y' as part of a trigonometric function (sin, cos, tan, etc.). To solve for 'y', apply trigonometric identities and properties to simplify the equation. Utilize inverse trigonometric functions or trigonometric identities to isolate 'y' and determine its value. Pay attention to the specific domain restrictions and periodicity of trigonometric functions.

Check Your Solution: After obtaining a value for 'y', it's crucial to check your solution by substituting it back into the original equation. Ensure that the equation holds true when 'y' is replaced with the determined value. This step confirms the accuracy of your solution and verifies that you haven't made any calculation errors.

Solving for 'y' is an essential skill in mathematics that allows you to find solutions to various types of equations. By understanding the equation type, isolating 'y', and employing appropriate techniques for linear, quadratic, exponential, logarithmic, and trigonometric equations, you can confidently find the value of 'y'. Regular practice and familiarity with algebraic operations will strengthen your ability to solve equations and deepen your mathematical understanding. Embrace the challenge, apply the techniques, and enjoy the satisfaction of successfully solving for 'y' in any equation that comes your way.

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allcalculator · 1 year

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Interesting queries on the Quadratic formula calculator!

What are the steps to use the quadratic equation calculator?

Allcalculator.net's The quadratic equation is a polynomial with the highest exponent, which should be 2. The standard form of the quadratic equation is Ax2 + Bx + C = 0.

Here are the steps to use the quadratic equation calculator are:

Enter the coefficients of the equation in the required input fields with the given statements.

Click on the button to solve the quadratic equation to get the exact result

In the final steps, you can find the result for the roots of the quadratic equation, which will be displayed in the output fields.

The basic quadratic formula is:

X = -b ± √b2 – 4ac / 2a

This above formula is used to solve all the quadratic equations where a ≠ 0 with a polynomial order of 2.

ax2 + bx + c = 0

Enter the equation for real and complex roots in the calculator to determine the discriminant (b2 – 4ac), which is less than, greater than or equal to 0.

When b2 – 4ac = 0, there is no real roots

When b2 – 4ac > 0, there are two real roots

When b2 – 4ac < 0, there are two complex roots

What are the types of quadratic equations?

There are three types of quadratic equations which are discussed below:

Standard form

It is a quadratic equation which is given by ax^2 + bx + c = 0

where a, b, and c are constants and x is a variable.

Vertex form

The Vertex form of a quadratic equation is given by a*(x - h) ^2 + k

A h and k are constants, and (h, k) is the vertex of the parabola

Factored form

The factored form of a quadratic equation is given by a*(x - r1) (x - r2)

Where a, r1, and r2 are constants and r1 and r2 are the roots or solutions of the quadratic equation

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longjoomla · 2 years

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Equation maker out of ordered pairs

EQUATION MAKER OUT OF ORDERED PAIRS FOR FREE

EQUATION MAKER OUT OF ORDERED PAIRS HOW TO

EQUATION MAKER OUT OF ORDERED PAIRS FREE

is a free online tool that displays the ordered pair for the given equation. The variable assigned to the rst component is also called the independent variable, and the variable assigned to the second component is called the dependent variable. Step 3: Click on 'Reset' to clear the field and enter new values. The ordered pairs (1,5),( 2, 1), and (12,4) all satisfy the equation y2 x+3. Step 2: Click on the 'Calculate' button to find the value of y with respect to x.

EQUATION MAKER OUT OF ORDERED PAIRS FOR FREE

The ordered pair ( 5, 2) works, since 2 5 3. Get the ordered pairs calculator available online for free only at BYJUS. Follow the steps given below to use the calculator: Step 1: Enter the equation and the one coordinate of the ordered pair in the space provided (enter the unknown as x). Example: Find an ordered pair which is a solution of the equation y x 3, and plot the point on the coordinate plane. and a given point in the plane located by using an ordered pair of numbers, called its coordinates. Where: a,b and c are the sides of the triangle. An ordered pair is often used to represent a point on a coordinate plane or the solution to an equation in two variables. Graphing Calculator+ is an iPhone/iPad calculator. S = semi perimeter of the triangle having this formula s = (a + b + c) / 2 Triangle perimeter formula Lets start by saying that a relation is simply a set or collection of ordered pairs. the side b expression is b = square root (c 2 - a 2) Heron formula for area of a triangleĪrea = square root (s(s - a)(s - b)(s - c)) the side a formula is a = square root (c 2 - b 2) This further confirms that the quadratic function is. among other notable applications, they are the fundamental tool in the. the hypotenuse the equation becomes c = square root (a 2 + b 2) As you can see, each horizontal line drawn through the graph of f(x) x2 passes through two ordered pairs. In mathematics, a partial differential equation (PDE) is an equation which imposes relations. Pythagorean theoremĪlso known as Pythagoras's theorem this states that in a right trianglethe square of the hypotenuse “c” (the side opposite the right angle) equals the sum of the squares of the other two sides “a” & “b”, thus its equation can be written as presented here: If the point is valid (a point is valid if the above condition is satisfied), increment the counter which stores the total number of valid points. Moreover it allows specifying angles either in grades or radians for a more flexibility. Method 1 (Brute Force): Generate all possible pairs (i, j) and check if a particular ordered pair (i, j) is such that, (arr i, arr j) satisfies the given equation of the line y mx + c, and i j. The algorithm of this right triangle calculator uses the Pythagorean theorem to calculate the hypotenuse or one of the other two sides, as well as the Heron formula to find the area, and the standard triangle perimeter formula as described below.

EQUATION MAKER OUT OF ORDERED PAIRS HOW TO

This tool is designed to find the sides, angles, area and perimeter of any right triangle if you input any 3 fields (any 3 combination between sides and angles) of the 5 sides and angles available in the form. How to calculate the equation of a linear function from two given points First, we have to calculate the slope m by inserting the x- and y- coordinates of the. How does this right triangle calculator work?

#Equation maker out of ordered pairs

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mademains · 2 years

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Quadratic inequalities

#QUADRATIC INEQUALITIES HOW TO#

To be neat, the smaller number should be on the left, and the larger on the right. The distance we want is from 10 m to 15 m: (Note: if you are curious about the formula, it is simplified from d = d 0 + v 0t + ½a 0t 2, where d 0=20 , We can use this formula for distance and time: We also know the endpoints are excluded since 3 creates a denominator of zero and we have a strict inequality.A stuntman will jump off a 20 m building.Ī high-speed camera is ready to film him between 15 m and 10 m above the ground. $$(-2)^2 - 5(-2) - 6 > 0$$ $$4 + 10 - 6 > 0$$ $$ equire$$ Since the test of the number 4 produces a false statement, we know values that are greater than 4 will not satisfy the inequality. Let's choose -2, we will plug this in for x in the original inequality. We can solve quadratic inequalities to give a range of. Let's begin with interval A, we can choose any value that is less than -1. Quadratic inequalities are similar to quadratic equations and when plotted they display a parabola. Step 3) Substitute a test number from each interval into the original inequality. This interval is labeled with the letter "C". The solution to a quadratic inequality in one variable can have no values, one value or. Let us consider the quadratic inequality x2 5x Lastly, we have an interval that consists of any number that is greater than 6. The standard quadratic equation becomes an inequality if it is represented as ax2 + bx + c 0).

This interval is labeled with the letter "B". One interval contains any number less than -1 and is labeled with the letter "A". Quadratic Inequalities Given 3 x 2 > -x + 4 Rewrite the inequality with one side equal to zero. On the horizontal number line, we can set up three intervals: We have split the number line up into three intervals. It shows the data which is not equal in graph form. Introduces a conceptual basis for solving quadratic inequalities, looking at linear inequalities and using a knowledge of what quadratic graphs look like. An equation is a statement that asserts the equality of two expressions and a linear inequality is an inequality which involves a linear function. These endpoints will allow us to set up intervals on the number line. In quadratic inequalities worksheets, we learn that a quadratic inequality is an equation of second degree that uses an inequality sign instead of an equal sign. Therefore, set the function equal to zero and solve. For a quadratic inequality in standard form, the critical numbers are the roots. $$x^2 - 5x - 6 > 0$$ Step 1) We will change this inequality into an equality and solve for x. It is important to note that this quadratic inequality is in standard form, with zero on one side of the inequality.

The endpoints are included for a non-strict inequality and excluded for a strict inequalityĮxample 1: Solve each inequality.

If a test number makes the inequality false, the region that includes that test number is not in the solution set.

If the test number makes the inequality true, the region that includes that test number is in the solution set.

If it is less than or greater than some number or any other polynomial. Consider a quadratic polynomial ax2+bx+c.

Substitute a test number from each interval into the original inequality Graphing a quadratic inequality is easier than you might think You just need to know the steps involved This tutorial takes you through those steps to. What do you mean by Quadratic inequalities.

For example, to solve a quadratic inequality -x2+x+2>0, we can find the values of x where the parabola.

Use the endpoints to set up intervals on the number line Students will solve quadratic inequalities and match each inequality with its solution set. Quadratic inequalities are best visualized in the plane.

These endpoints separate the solution regions from the non-solution regions.

If the quadratic inequality is not in one of the. is an example of a quadratic inequality, as it contains a single variable raised to the second power at maximum.

The solutions will give us the boundary points or endpoints Hence, we obtain four possible general forms of quadratic inequalities: ax 2 + bx + c > 0.

Replace the inequality symbol with an equality symbol and solve the equation.

Quadratic Inequalities A quadratic inequality is of the form: $$ax^2 + bx + c > 0$$ Where a ≠ 0, and our ">" can be replaced with any inequality symbol.

#QUADRATIC INEQUALITIES HOW TO#

In this lesson, we will learn how to solve quadratic and rational inequalities.

#Quadratic inequalities

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rewidumike · 2 years

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Canonical form of linear programming calculator

CANONICAL FORM OF LINEAR PROGRAMMING CALCULATOR >> Download vk.cc/c7jKeU

CANONICAL FORM OF LINEAR PROGRAMMING CALCULATOR >> Leia online bit.do/fSmfG

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The NumPy module is the de facto standard for numerical calculation in the Python programming language, a language whose use is growing rapidly in the Calibração de um modelo de regressão linear # Note que é possível usar simbolos como acca, A Canonical Correlation Analysis with Inferential Guaranties. Then we derive the algorithm for calculating the normal form of {\phi(0)} = c_2 {\Gamma(c_3)} $$ which if d is an integer can be expanded as a Pochammer linear, em que cada modo representa uma sequência de ações subsidiadas por técnicas especiais69. (TRLs) calculator software for systems engineering and. Solving Equations and Inequalities - Linear Equations, Quadratic Equations, Completing the Square, Quadratic Formula, Applications of Linear and Quadratic Linear programming e um substantivo. O nome ou substantivo é o tipo de palavras cujo significado determina a realidade. Os substantivos denominam todas as the problem in canonical form. Solve the first move from your starting location using the. Simplex method. Show your tableau for each step. Minimize :. canonical inflammasome activation dependent on caspase-11, gasdermin-D (GSDMD) and bacterial type IV secretion system. Additionally, we identify that Tomahawk -A new kind of music player that invites all your streams, downloads, Lightworks - Professional non-linear video editing program with a free

, , , , .

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kasey-writes-stuff · 3 years

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and since i spammed you, imma help you out (also you coulda googled these but you didnt and i respect that)

-the battle of agincourt was won by the english in 1415

-xenon is a noble gas

-correct; the middle phase is called interphase

-(-b±√(b^2-4ac))/(2a) is the quadratic formula and it's used to solve equations in standard form

-CORRECT nicely done it is from hamlet here have a cookie 🍪

-crème brûlée is indeed that thing you light on fire and then eat and as such, it translates into 'burnt cream'

-the saxophone is actually a woodwind instrument, which places it with the flutes and oboes, since it has a reed and not mouthpiece like other brass instruments

-haha you fell into my trap- i ask you a bunch of boring school questions and then you're caught off balance. thrown them off their rhythm

-those are super cute death spots btw

-aww lol

-and no you wont

love you!! /p

*evaporates*

Thank you for the answers and I did think of googling them but decided to be honest hdhdhd

Shut up they’re not cute especially since I’m not even sure they’re death spots smh /lh

You never know I might /lh

Love you too! /p

#asks #yo whose mans is this

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mathmadeeasy · 2 years

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What is Quadratic Equation? A quadratic condition is a logarithmic articulation of the second degree in x. The quadratic condition in its standard structure is ax2 + bx + c = 0, where a, b are the coefficients, x is the variable, and c is the steady term. The main condition for a situation to be a quadratic condition is the coefficient of x2 is a non-zero term(a ≠0). For composing a quadratic condition in standard structure, the x2 term is composed first, trailed by the x term, lastly, the steady term is composed. The numeric upsides of a, b, c are by and large not composed as parts or decimals yet are composed as vital qualities.

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Standard Form of a Quadratic Equation

Further in genuine numerical questions the quadratic conditions are introduced in various structures: (x - 1)(x + 2) = 0, - x2 = - 3x + 1, 5x(x + 3) = 12x, x3 = x(x2 + x - 3). These conditions should be changed into standard type of the quadratic condition prior to performing further tasks.

Quadratic Equation Quadratic conditions are second-degree logarithmic articulations and are of the structure ax2 + bx + c = 0. "Quadratic" is gotten from "Quad" and that implies square. As such, a quadratic condition is a "condition of degree 2." There are numerous situations where a quadratic condition is utilized. Do you have any idea that when a rocket is sent off, its way is depicted by a quadratic condition? Further, a quadratic condition has various applications in physical science, designing, cosmology.

The quadratic conditions are second-degree conditions in x that have two solutions for x. These two responses for x are likewise called the foundations of the quadratic conditions and are assigned as (α, β). We will study the foundations of a quadratic condition down below.

Solve quadratic equation in 10seconds👇

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Standard Form of a Quadratic Equation

Quadratic Equation Formula Quadratic Formula is the easiest method for tracking down the foundations of a quadratic condition. There are sure quadratic conditions that won't be quickly factorized, and here we can helpfully utilize this quadratic recipe to track down the roots in the fastest manner. The foundations of the quadratic condition further assistance to track down the amount of the roots and the result of the foundations of the quadratic condition. The two roots in the quadratic recipe are introduced as a solitary articulation. The positive sign and the negative sign can be on the other hand used to get the two particular foundations of the situation.

Quadratic Formula = [-b ± √(b² - 4ac)]/2a

Quadratic Formula

Significant Formulas for Solving Quadratic Equations The accompanying rundown of significant recipes is useful to tackle quadratic conditions.

The quadratic condition in its standard structure is ax2 + bx + c = 0 The discriminant of the quadratic condition is D = b2 - 4ac For D > 0 the roots are genuine and particular. For D = 0 the roots are genuine and equivalent. For D < 0 the roots don't exist, or the roots are fanciful. The recipe to observe the foundations of the quadratic condition is x = − b ± √ b 2 − 4 a c 2 a . The amount of the foundations of a quadratic condition is α + β = - b/a = - Coefficient of x/Coefficient of x2. The result of the Root of the quadratic condition is αβ = c/a = Constant term/Coefficient of x2 The quadratic condition having roots α, β, is x2 - (α + β)x + αβ = 0. The condition for the quadratic conditions a 1 x 2 + b 1 x + c

0 , also, a 2 x 2 + b 2 x + c

0 having similar roots is ( a 1 b 2 − a 2 b 1 ) ( b 1 c 2 − b 2 c 1 ) = ( a 2 c 1 − a 1 c 2 ) 2 . For positive upsides of (a > 0), the quadratic articulation f(x) = ax2 + bx + c has a base worth at x = - b/2a. For negative worth of (a < 0), the quadratic articulation f(x) = ax2 + bx + c has a greatest worth at x = - b/2a. For a > 0, the scope of the quadratic condition ax2 + bx + c = 0 is [b2 - 4ac/4a, ∞) For a < 0, the scope of the quadratic condition ax2 + bx + c = 0 is : (∞, - (b2 - 4ac)/4a] Quadratic Formula Proof Think about an erratic quadratic condition: ax2 + bx + c = 0, a ≠ 0

To decide the foundations of this situation, we continue as follows:

ax2 + bx = - c ⇒ x2 + bx/a = - c/a

Presently, we express the left hand side as an ideal square, by presenting another term (b/2a)2 on the two sides:

x2+ bx/a + (b/2a)2 = - c/a + (b/2a)2

The left hand side is presently an ideal square:

(x + b/2a)2 = - c/a + b2/4a2 ⇒ (x + b/2a)2 = (b2 - 4ac)/4a2

This is great as far as we're concerned, on the grounds that now we can take square roots to get:

x + b/2a = ±√(b2 - 4ac)/2a

x = (- b ± √(b2 - 4ac))/2a

Hence, by finishing the squares, we had the option to detach x and get the two foundations of the situation.

Foundations of a Quadratic Equation The foundations of a quadratic condition are the two upsides of x, which are gotten by settling the quadratic condition. The foundations of a quadratic condition are alluded to by the images alpha (α), and beta (β). These foundations of the quadratic condition are likewise called the zeros of the situation. Here we will study how to track down the idea of foundations of a quadratic condition without really tracking down the foundations of the situation. And furthermore look at the recipes to track down the total and the result of the foundations of the situation.

Nature of Roots of the Quadratic Equation The idea of foundations of a quadratic condition can be found without really tracking down the roots (α, β) of the situation. This is conceivable by taking the discriminant esteem, which is important for the recipe to settle the quadratic condition. The worth b2 - 4ac is known as the discriminant of a quadratic condition, and is assigned as 'D'. In view of the discriminant esteem the idea of the foundations of the quadratic condition can be anticipated.

Discriminant: D = b2 - 4ac D > 0, the roots are genuine and particular D = 0, the roots are genuine and equivalent. D < 0, the roots don't exist or the roots are fanciful. Nature of Roots of a Quadratic Equation

Connection Between Coefficients and Roots of Quadratic Equation The coefficient of x2, x term, and the steady term of the quadratic condition ax2 + bx + c = 0 are helpful to concentrate on additional about the properties of foundations of the quadratic condition. The total and result of foundations of a quadratic condition can be straightforwardly determined from the situation, without really tracking down the foundations of the quadratic condition. The amount of the foundations of the quadratic condition is equivalent to the negative of the coefficient of x isolated by the coefficient of x2. The result of the base of the situation is equivalent to the steady term isolated by the coefficient of the x2. For a quadratic condition ax2 + bx + c = 0, the total and result of the roots are as per the following.

Amount of the Roots: α + β = - b/a = - Coefficient of x/Coefficient of x2 Result of the Roots: αβ = c/a = Constant term/Coefficient of x2 The quadratic condition can likewise be shaped for the given foundations of the situation. In the event that α, β, are the foundations of the quadratic condition, the quadratic condition is as per the following.

x2 - (α + β)x + αβ = 0

Techniques to Solve Quadratic Equations A quadratic condition can be tackled to get two upsides of x or the two foundations of the situation. There are four unique strategies to track down the foundations of the quadratic condition. The four techniques for tackling the quadratic conditions are as per the following.

Factorizing of Quadratic Equation Recipe Method of Finding Roots Strategy for Completing the Square Charting Method to Find the Roots Allow us to examine detail at every one of the above strategies to comprehend how to utilize these techniques, their applications, and their purposes.

Factorization of Quadratic Equation Factorization of quadratic condition follows a grouping of steps. For an overall type of the quadratic condition ax2 + bx + c = 0, we want to initially part the center term into two terms, to such an extent that the result of the terms is equivalent to the steady term. Further, we can take the normal terms from the accessible term, to get the expected factors at last. For understanding factorization, the overall type of the quadratic condition can be introduced as follows.

x2 + (a + b)x + stomach muscle = 0 x2 + hatchet + bx + stomach muscle = 0 x(x + a) + b(x + a) (x + a)(x + b) = 0 Allow us to comprehend factorization through the beneath model.

x2 + 5x + 6 = 0 x2 + 2x + 3x + 6 = 0 x(x + 2) + 3(x + 2) = 0 (x + 2)(x + 3) = 0 Hence the two got elements of the quadratic condition are (x + 2) and (x + 3).

Quadratic Formula to Find Roots The quadratic conditions which can't be tackled through the strategy for factorization can be settled with the assistance of a recipe. The recipe to tackle the quadratic condition utilizes the terms from the standard type of a quadratic condition. Through the beneath recipe we can get the two foundations of x by first involving the positive sign in the equation and afterward utilizing the negative sign. Any quadratic condition can be settled utilizing this recipe.

Quadratic Equation Formula

Further to the previously mentioned two strategies for tackling quadratic conditions, there is one more significant strategy for settling a quadratic condition. The strategy for finishing the square for a quadratic condition is likewise helpful to track down the foundations of the situation. This strategy incorporates various logarithmic estimations and thus has been made sense of as a different theme.

Strategy for Completing the Square The strategy for finishing the square for a quadratic condition, is to mathematically square and improve, to get the expected foundations of the situation. Consider a quadratic condition ax2 + bx + c = 0, a ≠ 0. To decide the foundations of this situation, we work on it as follows:

ax2 + bx + c = 0 ax2 + bx = - c x2 + bx/a = - c/a Presently, we express the left hand side as an ideal square, by presenting another term (b/2a)2 on the two sides:

x2 + bx/a + (b/2a)2 = - c/a + (b/2a)2 (x + b/2a)2 = - c/a + b2/4a2 (x + b/2a)2 = (b2 - 4ac)/4a2 x + b/2a = +√(b2-4ac)/2a Presently with this strategy for finishing the square, we could merge the incentive for the foundations of the situation. Further on improvement and taking the square root, the two potential foundations of the quadratic condition are, x = (- b + √(b2-4ac))/2a. Here the '+' sign gives one root and the '- ' sign gives one more base of the quadratic condition.

Quadratic Formula = [-b ± √(b² - 4ac)]/2a

Quadratic Formula

Significant Formulas for Solving Quadratic Equations The accompanying rundown of significant recipes is useful to tackle quadratic conditions.

0 , also, a 2 x 2 + b 2 x + c

To decide the foundations of this situation, we continue as follows:

ax2 + bx = - c ⇒ x2 + bx/a = - c/a

Presently, we express the left hand side as an ideal square, by presenting another term (b/2a)2 on the two sides:

x2+ bx/a + (b/2a)2 = - c/a + (b/2a)2

The left hand side is presently an ideal square:

(x + b/2a)2 = - c/a + b2/4a2 ⇒ (x + b/2a)2 = (b2 - 4ac)/4a2

This is great as far as we're concerned, on the grounds that now we can take square roots to get:

x + b/2a = ±√(b2 - 4ac)/2a

x = (- b ± √(b2 - 4ac))/2a

Hence, by finishing the squares, we had the option to detach x and get the two foundations of the situation.

x2 - (α + β)x + αβ = 0

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mathsai · 9 months

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How to Solve Quadratic Equations: Maths.ai

Quadratic equations are an essential part of algebra, and they often pose challenges for students when it comes to solving them. However, with the advent of artificial intelligence, learning and understanding complex maths problems has become more accessible than ever. Maths.ai, an AI-based website dedicated to assisting students in solving maths equations, introduces a powerful chatbot capable of guiding users through the process of solving quadratic equations. In this article, we will explore how to solve quadratic equations using the Maths.ai chatbot. It provides a comprehensive step-by-step guide to help students grasp this topic more effectively.

Understanding Quadratic Equations

Before diving into the solution process, it is essential to understand what a quadratic equation is and how it is represented. A quadratic equation is a second-degree polynomial equation in a single variable (usually denoted by "x"). It takes the general form:

ax^2 + bx + c = 0

where "a," "b," and "c" are constants, and "a" must not be equal to zero.

Introducing the Maths.ai Chatbot

Maths.ai is an intelligent virtual assistant designed to offer step-by-step guidance on various maths problems, including quadratic equations. By utilizing natural language processing and machine learning algorithms, the chatbot can understand user queries and provide personalized solutions, making it an invaluable tool for students seeking to improve their maths skills.

Step-by-Step Guide to Solving Quadratic Equations with Maths.ai

Now, let's delve into the process of solving quadratic equations using the Maths.AI chatbot. Here's a comprehensive step-by-step guide:

Step 1: Accessing the Chatbot

To get started, students can visit the Maths.ai website and locate the chatbot interface. Upon accessing it, the chatbot will greet the user and prompt them to enter the quadratic equation they wish to solve.

Step 2: Entering the Quadratic Equation

The user can input the quadratic equation in the standard form (ax^2 + bx + c = 0) directly into the chatbot's text box. The AI-powered chatbot will then analyze the equation and check for any errors in the input.

Step 3: Validating the Equation

After receiving the quadratic equation, the chatbot will verify if it is indeed a quadratic equation by checking that "a" is not equal to zero. If the equation is not in the proper form or violates any fundamental rules, the chatbot will provide feedback to the user, asking them to re-enter the equation correctly.

Step 4: Identifying the Coefficients

Once the chatbot validates the equation, it will identify and extract the values of "a," "b," and "c" from the entered quadratic equation.

Step 5: Computing the Discriminant

The chatbot will calculate the discriminant based on the coefficients "a," "b," and "c."

Step 6: Analyzing the Nature of Roots

The value of the discriminant (Δ) provides insight into the nature of the roots of the quadratic equation:The discriminant (Δ) is a crucial value in determining the nature of the roots of a quadratic equation. It is computed as follows:

Δ = b^2 - 4ac

a. If Δ > 0, the equation has two distinct real roots.

b. If Δ = 0, the equation has one real root (a repeated root).

c. If Δ < 0, the equation has two complex (non-real) roots.

The chatbot will inform the user about the nature of the roots based on the calculated discriminant.

Step 7: Calculating the Roots

Next, the chatbot will guide the user through the process of finding the roots of the quadratic equation based on the nature of the roots:

a. If the equation has two distinct real roots, the roots can be computed using the quadratic formula:

x = (-b ± √Δ) / 2a

b. If the equation has one real root (a repeated root), the root can be calculated as:

x = -b / 2a

c. If the equation has two complex (non-real) roots, the roots can be expressed in the form:

x = (-b ± √|Δ|i) / 2a

where "i" represents the imaginary unit (i.e., √-1).

Step 8: Presenting the Solutions

Finally, the chatbot will display the solutions of the quadratic equation to the user. It will provide both the exact values (in case of real roots) and the expressions (in case of complex roots) for better understanding.

In conclusion, quadratic equations are a fundamental concept in algebra, and solving them can be simplified with the assistance of Maths.ai. By following the step-by-step guide outlined in this article, students can effectively solve quadratic equations and gain a deeper understanding of the process. Maths.ai's AI-based chatbot is revolutionizing the way students approach math problems, making learning more engaging and accessible than ever before. As technology continues to advance, we can expect further innovations like the Maths.ai to transform the landscape of education.

#mathsai #mathematics

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tinytragedynacho · 3 years

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Ncert solutions for class 3

Ncert solutions for class 3 English comprehensions are similar to Ncert solutions but differ in a few areas. The main difference between these two exams is that the former tests a person's understanding of the meaning of the terms used in the language. Whereas the latter focuses more on reading and writing skills. It is therefore important that when you prepare for these exams you consider how much you've progressed in both areas before looking at your English grammar scores.

Most NCERT textbooks cover issues from the current editions of all NCERT Class 3 Hindi texts. This includes the explanations of each topic and their corresponding practice examinations. The reality that Class 3 is such an important milestone in your educational career just adds to why you must pay special attention to every subject in your syllabus. In this guide, you will get tips and advice on preparing for the exams for each of the three classes included in the NCERT standard curriculum. Some of the topics you will study are as follows:

o Test preparation: When you choose to go through the reviews for the NCERT exam you will be asked to do several practice tests based on previously prepared material. Some of these review papers include questions based on a mix of previous exam papers and practice materials. As part of your preparation for the NCERT solutions for class 3 English comprehensions, you should also be preparing for a mock exam based on a sample exam. This will allow you to see how you are fair against the sample exam, as well as getting an idea of the kinds of questions you will face on the actual exam.

o Class 9 English fluency: When it comes to the NCERT exam, reading is not the only thing we learn. Not only do we have to be able to understand what we read, but we also have to be able to understand how we read it. In this class, you will learn about the different forms of literary language including the key constituents and differences between the various genres.

o Ncert solutions for class 4 maths: When you take the NCERT tests, you will be required to demonstrate your knowledge of arithmetic. This will involve answering a set of questions related to the subject matter, normally based on weekly issues. As part of your preparation, you will be expected to work through a range of practice questions that test both your understanding and basic skills. These will help you to build up what is known as speed reading' so that you can do well on the actual exam. One of the key things to remember about answering NCERT questions is to begin and finish answering them at the same time, completing your response before the other answers are displayed is one way of achieving this.

o Ncert solutions for class 8 maths: The syllabus for this test includes some topics that are covered not only in the main body of the course but in many of the sub-areas as well. One of these areas is Ncert solutions for class 8 maths which will include working through the quadratic equation and solving for the roots of the cubic boron polynomial. This is a key topic as it will help students to understand the inner workings of this integral and how to solve it for the purpose of finding out the answer using the right method. Another key area of this course is the quadratic formula which is used extensively in science and engineering and often found to be written off by students without being able to explain it in an understandable way. A good number of students will therefore look towards the use of a pre-algated solution for the problems when they cannot find an easy and clear answer on their own.

Ncert solutions for class 9 English. The syllabus for this test has some very standard Ncert solutions for class 9 English which will cover the basics of sentence structure, how to introduce a topic, and even how to give an explanation about a topic. The main topics covered are subject pronouns, demonstrative pronouns, question words, and prepositions. An important thing to remember when looking at the syllabus is that in every lesson there will be at least two or three topics that will be covered which will differ slightly from the rest of the lessons.

o Ncert solutions for class 6 science. The syllabus for this test includes quite a few standard Ncert solutions for class 6 science which are designed to test students’ understanding of properties of matter. These include topics such as atoms, molecules, matter, solid-state structures, electronic and optical properties, electromagnetic properties, and nuclear properties. Being able to understand the concepts behind each of these will allow students to grasp the material and therefore improve their grades.

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educationtech · 3 years

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How to Prepare Maths for JEE Mains 2021 - Arya College

Mathematics is the most essential parts of the entrance. In fact, it is an indispensable part of human race. Chemistry, Physics and Mathematics have an equal weightage in the JEE Mains but Mathematics fulfils an edge over the other two. The score in mathematics is always considered to be the deciding factor. Maths is the king of science and the base for Physics and Chemistry. Topics like differential equations, calculus and quadratic equations are used in majority questions of physics and chemistry.

JEE Mathematics is easy to learn and understand but students fail to excel in it as their basics are not very clear. Some of the topics are easy to crack in the examination like permutations and combinations, trigonometry, quadratic equations, probability, straight lines and circles in coordinate geometry and differential calculus. Hard work and practice are the only way to crack the JEE examination for the students of BTech Colleges Jaipur. The MCQ format only test the applicant’s knowledge base and their aptitude.

The most significant thing while tackling a competitive examination is having clear basics. A lot of math books are present that explains all the essential concepts in detail and promise to hold the most difficult form of questions, but what matters is how the basics are comprehended. There are some of the important topics from where most of the questions in JEE Mains 2021 are generally asked.

Topics to cover in JEE Math Section1. Algebra

No other topic can come close to Algebra in the simplicity and ability to set framework for easily obtainable marks in the JEE Mains Examination. The applicant is required to have high computational speed and clarity about their basic concepts. Also, algebra is considered as one of the most interesting topics. Thus, the clarity of concepts plus good calculation speed is enough to score outstanding marks from this topic. Some of the major areas that an applicant should cover in algebra includes Differential Equations, Probability, Permutation and Combination, Complex Numbers, Quadratic Equations and Expressions, Binomial Theorem, Matrices and Determinants, Progressions and Series, Sets, Relations, and Functions, etc.

2. Calculus

It is among the largest section of modern mathematics and is considered as one of the important topics for JEE Mains preparation for the students of Private Engineering Colleges Jaipur. Calculus has two main sections. Differential and Integral Calculus, related with the fundamental theory of calculus. It deals with studying operations and applying it to solve the equations. Problems with a base in theory as well as problems that have their base in application show equal appearance in JEE Mains Maths section. The differential calculus includes limits and Continuity, Application of Derivatives, Differentiation and integral calculus includes indefinite Integral, Definite Integral, Area under Curves, etc.

3. Coordinate Geometry

One other essential constituent of the math section in JEE Mains Exams is coordinate geometry. Other entrance examinations consider it as one of the most scorable subjects and if explained in the most basic sense, it is the analyzation of geometrical shape. Also, this is one of the only topics in which students of best engineering colleges Jaipur can secure full marks. Some of the major sections of it are Straight Lines, Parabola, Hyperbola, Circle, Ellipse, etc.

Most of the math problems are frequently created using the topics of Circle and Straight line. These sections when combined with each other contribute a lot of questions to the JEE examination and are a prerequisite to conic sections as well.

4.Trigonometry

An essential branch of maths in the JEE exam is Trigonometry. Some of important areas of trigonometry are trigonometric ratios and trigonometric functions. Applicants of best btech colleges Jaipur should be extremely clear about their fundamentals of trigonometry, as many competitive exams ask numerous questions from this topic to clear the examination. Sides and angles of a triangle are associated and this relationship is studied under the branch of trigonometry. A lot of branches of technology and science have their bases in the functions of trigonometry. The JEE Trigonometry problems ranges from the trigonometry basics to the applications of trigonometry.

General Math study tips for JEE1. Bring a change in your mindset

The first thing that students of the best BTech Colleges should do is develop a positive attitude towards the subject. Maths is not a subject to afraid. Be confident and ready to learn new things.

2. Start your preparations early on

Students of Engg colleges should start their preparations for JEE from an early stage. Usually, students start from 11th standard itself. During this, you understand the JEE Math syllabus for JEE Advanced and JEE Mains exam and understand the learning objectives of the question setters. Find out about the significant topics and create a preparation strategy accordingly.

3. Practice till the concepts become familiar

During this stage, students of engineering colleges can start by learning the basic concepts and then move on to more advanced ones like hyperbola or binomial theorem. If an individual want to master Maths concepts, they cannot do it by just reading about the concepts or steps on how to solve the problems. They should start solving problems until they fully understand the core idea of the concept. At the same time, they should manage time properly while doing it.

4. Don’t engage in rote learning

Memorizing concepts or rote learning should be avoided if students of top Engg colleges are dealing with Maths. This subject requires more practical and reasoning skills. It is significant for you to understand the concepts clearly and also have answers for both the “How” as well as the “Why” part.

5. Create a maths dictionary

While Maths deals with a lot of formulas, have a separate notebook for them. Students of top BTech colleges Rajasthan can write all of them in this notebook and have a look at them at a later instance. Also, they can make or keep a maths dictionary, flashcards that can help them to quickly and easily revise or remember the formulas and at any moment.

Conclusion

Students of the list of engineering colleges in Jaipur can score good and achieve great rankings in the examination by preparing these subjects proficiently. Every question has a short and sweet method of being solved. Thus, applicants gain command over the different topics as they surf through them regularly. An individual should practice previous years JEE Main question papers as it would give them an idea of how the paper is set and what number of marks are allotted to what topic.

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dogshunter345 · 3 years

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FX Equation

How To Solve Fx Equations

Fx-991ms Equation Solver

Fx Equation

Fx Equations

FxSolver is a math solver for engineering and scientific equations. To get started, add some formulas, fill in any input variables and press 'Solve.' Solver Browse formulas Create formulas new Sign in. A Quick Look at FX Equation 5. Enter an equation into the text box at the left by clicking in the palette or typing with the keyboard. Note: to use Greek symbols such as θ, type out the letter's name in English, e.g.

Step 1 : Enter a quadratic function in terms of x.

Step 2 : Choose a command relating to the function f(x) you entered above.

Rewrite f(x) in the form a(x ± b)2 ± c by completing the square.

For the parabola y = f(x), calculate the following :

In finance, a foreign exchange option (commonly shortened to just FX option or currency option) is a derivative financial instrument that gives the right but not the obligation to exchange money denominated in one currency into another currency at a pre-agreed exchange rate on a specified date. FX Equation is simply the fastest way to produce mathematical and scientific equations. It provides an equation creation environment that really is 'Astoundingly Quick'. It adds automatic vertical alignment of equals signs and a high speed toolbar entry system that can speed up entry of more complicated equation.

Solving equations is the central theme of algebra. All skills learned lead eventually to the ability to solve equations and simplify the solutions. In previous chapters we have solved equations of the first degree. You now have the necessary skills to solve equations of the second degree, which are known as quadratic equations.

QUADRATICS SOLVED BY FACTORING

OBJECTIVES

Upon completing this section you should be able to:

Identify a quadratic equation.

Place a quadratic equation in standard form.

Solve a quadratic equation by factoring.

A quadratic equation is a polynomial equation that contains the second degree, but no higher degree, of the variable.

The standard form of a quadratic equation is ax2 + bx + c = 0 when a ≠ 0 and a, b, and c are real numbers.

All quadratic equations can be put in standard form, and any equation that can be put in standard form is a quadratic equation. In other words, the standard form represents all quadratic equations.

The solution to an equation is sometimes referred to as the root of the equation.

This theorem is proved in most college algebra books.

An important theorem, which cannot be proved at the level of this text, states 'Every polynomial equation of degree n has exactly n roots.' Using this fact tells us that quadratic equations will always have two solutions. It is possible that the two solutions are equal.

A quadratic equation will have two solutions because it is of degree two.

The simplest method of solving quadratics is by factoring. This method cannot always be used, because not all polynomials are factorable, but it is used whenever factoring is possible.

The method of solving by factoring is based on a simple theorem.

If AB = 0, then either A = 0 or B = 0.

In other words, if the product of two factors is zero, then at least one of the factors is zero.

We will not attempt to prove this theorem but note carefully what it states. We can never multiply two numbers and obtain an answer of zero unless at least one of the numbers is zero. Of course, both of the numbers can be zero since (0)(0) = 0.

SolutionStep 1 Put the equation in standard form.

We must subtract 6 from both sides.

Step 2 Factor completely.

Recall how to factor trinomials.

Step 3 Set each factor equal to zero and solve for x. Since we have (x - 6)(x + 1) = 0, we know that x - 6 = 0 or x + 1 = 0, in which case x = 6 or x = - 1.

This applies the above theorem, which says that at least one of the factors must have a value of zero.

Step 4 Check the solution in the original equation. If x = 6, then x2 - 5x = 6 becomes

Checking your solutions is a sure way to tell if you have solved the equation correctly.

Therefore, x = 6 is a solution. If x = - 1, then x2 - 5x = 6 becomes

Therefore, - 1 is a solution.

The solutions can be indicated either by writing x = 6 and x = - 1 or by using set notation and writing {6, - 1}, which we read 'the solution set for x is 6 and - 1.' In this text we will use set notation.

In this example 6 and -1 are called the elements of the set.

Note in this example that the equation is already in standard form.

Again, checking the solutions will assure you that you did not make an error in solving the equation. are also called roots of the equation.

(x + 1) is the least common denominator of all the fractions in the equation. Remember, every term of the equation must be multiplied by (x + 1).

Check the solutions in the original equation.

Check in the original equation to make sure you do not obtain a denominator with a value of zero.

Notice here the two solutions are equal. This only occurs when the trinomial is a perfect square.

INCOMPLETE QUADRATICS

OBJECTIVES

Upon completing this section you should be able to:

Identify an incomplete quadratic equation.

Solve an incomplete quadratic equation.

If, when an equation is placed in standard form ax2 + bx + c = 0, either b = 0 or c = 0, the equation is an incomplete quadratic.

Example 1

5x2 - 10 = 0 is an incomplete quadratic, since the middle term is missing and therefore b = 0.

When you encounter an incomplete quadratic with c - 0 (third term missing), it can still be solved by factoring.

x is a common factor. The product of two factors is zero. We therefore use the theorem from the previous section. Check these solutions.

Notice that if the c term is missing, you can always factor x from the other terms. This means that in all such equations, zero will be one of the solutions. An incomplete quadratic with the b term missing must be solved by another method, since factoring will be possible only in special cases.

Example 3 Solve for x if x2 - 12 = 0.

Solution Since x2 - 12 has no common factor and is not the difference of squares, it cannot be factored into rational factors. But, from previous observations, we have the following theorem.

Note that there are two values that when squared will equal A.

Using this theorem, we have

Check these solutions.

Add 10 to each side. Check these solutions.

Here 7x is a common factor. Check these solutions.

Note that in this example we have the square of a number equal to a negative number. This can never be true in the real number system and, therefore, we have no real solution.

COMPLETING THE SQUARE

OBJECTIVES

Upon completing this section you should be able to:

Identify a perfect square trinomial.

Complete the third term to make a perfect square trinomial.

Solve a quadratic equation by completing the square.

From your experience in factoring you already realize that not all polynomials are factorable. Therefore, we need a method for solving quadratics that are not factorable. The method needed is called 'completing the square.'

First let us review the meaning of 'perfect square trinomial.' When we square a binomial we obtain a perfect square trinomial. The general form is (a + b)2 = a2 + 2ab + b2.

Remember, squaring a binomial means multiplying it by itself.

From the general form and these examples we can make the following observations concerning a perfect square trinomial.

Two of the three terms are perfect squares. 4x2 and 9 in the first example, 25x2 and 16 in the second example, and a2 and b2 in the general form.

In other words, the first and third terms are perfect squares.

The other term is either plus or minus two times the product of the square roots of the other two terms.

The -7 term immediately says this cannot be a perfect square trinomial. The task in completing the square is to find a number to replace the -7 such that there will be a perfect square.

Consider this problem: Fill in the blank so that 'x2 + 6x + _______' will be a perfect square trinomial. From the two conditions for a perfect square trinomial we know that the blank must contain a perfect square and that 6x must be twice the product of the square root of x2 and the number in the blank. Since x is already present in 6x and is a square root of x2, then 6 must be twice the square root of the number we place in the blank. In other words, if we first take half of 6 and then square that result, we will obtain the necessary number for the blank.

Therefore x2 + 6x + 9 is a perfect square trinomial.

Now let's consider how we can use completing the square to solve quadratic equations.

Example 5 Solve x2 + 6x - 7 = 0 by completing the square.

Recall that instead of -7, a +9 would make the expression a perfect square.

Solution First we notice that the -7 term must be replaced if we are to have a perfect square trinomial, so we will rewrite the equation, leaving a blank for the needed number.

At this point, be careful not to violate any rules of algebra. For instance, note that the second form came from adding +7 to both sides of the equation. Never add something to one side without adding the same thing to the other side.

Now we find half of 6 = 3 and 32 = 9, to give us the number for the blank. Again, if we place a 9 in the blank we must also add 9 to the right side as well.

Remember, if 9 is added to the left side of the equation, it must also be added to the right side.

Now factor the perfect square trinomial, which gives

Now x2 + 6x + 9 may be written as (x + 3)2.

Add - 3 to both sides.

Thus, 1 and -7 are solutions or roots of the equation.

Example 6 Solve 2x2 + 12x - 4 = 0 by completing the square.

Solution This problem brings in another difficulty. The first term, 2x2, is not a perfect square. We will correct this by dividing all terms of the equation by 2 and obtain

In other words, obtain a coefficient of 1 for the x2 term.

We now add 2 to both sides, giving

Again, this is more concise.

Example 7 Solve 3x2 + 7x - 9 = 0 by completing the square.

SolutionStep 1 Divide all terms by 3.

Again, obtain a coefficient of 1 for x2 by dividing by 3.

Step 2 Rewrite the equation, leaving a blank for the term necessary to complete the square.

Step 3 Find the square of half of the coefficient of x and add to both sides.

It looks complex, but we are following the same exact rules as before.

Step 4 Factor the completed square.

The factoring should never be a problem since we know we have a perfect square trinomial, which means we find the square roots of the first and third terms and use the sign of the middle term. You should review the arithmetic involved in adding the numbers on the right at this time if you have any difficulty. We now have

Step 5 Take the square root of each side of the equation.

Step 6 Solve for x (two values).

cannot be simplified. We could also write the solution to this problem in a more condensed form as

Follow the steps in the previous computation and then note especially the last ine. What is the conclusion when the square of a quantity is equal to a negative number? 'No real solution.'

What real number can we square and obtain -7?

In summary, to solve a quadratic equation by completing the square, follow this step-by-step method.

Step 1 If the coefficient of x2 is not 1, divide all terms by that coefficient. Step 2 Rewrite the equation in the form of x2 + bx + _______ = c + _______. Step 3 Find the square of one-half of the coefficient of the x term and add this quantity to both sides of the equation. Step 4 Factor the completed square and combine the numbers on the right-hand side of the equation. Step 5 Find the square root of each side of the equation. Step 6 Solve for x and simplify. If step 5 is not possible, then the equation has no real solution.

These steps will help in solving the equations in the following exercise.

THE QUADRATIC FORMULA

OBJECTIVES

Upon completing this section you should be able to:

Solve the general quadratic equation by completing the square.

Solve any quadratic equation by using the quadratic formula.

Solve a quadratic equation by completing the square.

The standard form of a quadratic equation is ax2 + bx + c = 0. This means that every quadratic equation can be put in this form. In a sense then ax2 + bx + c = 0 represents all quadratics. If you can solve this equation, you will have the solution to all quadratic equations.

We will solve the general quadratic equation by the method of completing the square.

This is to obtain an x2 term with a coefficient of 1. This we did in the previous section many times.

We must add to each side.

This form is called the quadratic formula and represents the solution to all quadratic equations.

Memorize this expression.

To use the quadratic formula you must identify a, b, and c. To do this the given equation must always be placed in standard form.

Carefully substitute the values of a, b, and c in the formula.

Not every quadratic equation will have a real solution.

This equation is already in standard form.

There is no real solution since -47 has no real square root.

Again, this equation is in standard form.

This solution should now be simplified.

WORD PROBLEMS

How To Solve Fx Equations

OBJECTIVES

Upon completing this section you should be able to:

Identify word problems that require a quadratic equation for their solution.

Solve word problems involving quadratic equations.

Certain types of word problems can be solved by quadratic equations. The process of outlining and setting up the problem is the same as taught in chapter 5, but with problems solved by quadratics you must be very careful to check the solutions in the problem itself. The physical restrictions within the problem can eliminate one or both of the solutions.

Example 1 If the length of a rectangle is 1 unit more than twice the width, and the area is 55 square units, find the length and width.

Solution The formula for the area of a rectangle is Area = Length X Width. Let x = width, 2x + 1 = length.

If x represents the width, then 2x represents twice the width and 2x + 1 represents one more than twice the width.

Place the quadratic equation in standard form. This quadratic can be solved by factoring.

At this point, you can see that the solution x = -11/2 is not valid since x represents a measurement of the width and negative numbers are not used for such measurements. Therefore, the solution is

width = x = 5, length = 2x + 1 = 11.

A measurement cannot be a negative value.

The reciprocal of x is . Remember LCD means lowest common denominator. Every term must be multiplied by 10x. Again, this quadratic can be factored.

Both solutions check. Therefore, the solution set is .

There are two solutions to this problem.

Example 3 If a certain integer is subtracted from 6 times its square, the result is 15. Find the integer.

Solution Let x = the integer. Then

Since neither solution is an integer, the problem has no solution.

You might be tempted to give these values as a solution unless you paid close attention to the fact that the problem asked for an integer.

Example 4 A farm manager has 200 meters of fence on hand and wishes to enclose a rectangular field so that it will contain 2,400 square meters in area. What should the dimensions of the field be?

Solution Here there are two formulas involved. P = 2l + 2w for the perimeter and A = lw for the area. First using P = 2l + 2w, we get

Divide each term by 2.

We can now use the formula A = lw and substitute (100 - l) for w, giving

The field must be 40 meters wide by 60 meters long.

We could just as well solve for l obtaining l = 100 - w. Then

Fx-991ms Equation Solver

Note that in this problem we actually use a system of equations

Fx Equation

P = 2 l + 2 w A = l w.

In general, a system of equations in which a quadratic is involved will be solved by the substitution method. (See chapter 6.)

SUMMARY

Key Words

A quadratic equation is a polynomial equation in one unknown that contains the second degree, but no higher degree, of the variable.

The standard form of a quadratic equation is ax2 + bx + c = 0, when a ≠ 0.

An incomplete quadratic equation is of the form ax2 + bx + c = 0, and either b = 0 or c = 0.

The quadratic formula is

Fx Equations

Procedures

The most direct and generally easiest method of finding the solutions to a quadratic equation is factoring. This method is based on the theorem: if AB = 0, then A = 0 or B = 0. To use this theorem we put the equation in standard form, factor, and set each factor equal to zero.

To solve a quadratic equation by completing the square, follow these steps: Step 1 If the coefficient of x2 is not 1, divide all terms by that coefficient. Step 2 Rewrite the equation in the form of x2 + bx +_____ = c + _____ Step 3 Find the square of one-half of the coefficient of the x term and add this quantity to both sides of the equation. Step 4 Factor the completed square and combine the numbers on the right-hand side of the equation. Step 5 Find the square root of each side of the equation. Step 6 Solve for x and simplify.

The method of completing the square is used to derive the quadratic formula.

To use the quadratic formula write the equation in standard form, identify a, b, and c, and substitute these values into the formula. All solutions should be simplified.

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