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• Mathematics ## How to Solve Quadratic Equations

Last Updated: February 10, 2023 Fact Checked

A quadratic equation is a polynomial equation in a single variable where the highest exponent of the variable is 2.  X Research source There are three main ways to solve quadratic equations: 1) to factor the quadratic equation if you can do so, 2) to use the quadratic formula, or 3) to complete the square. If you want to know how to master these three methods, just follow these steps.

## Factoring the Equation • Then, use the process of elimination to plug in the factors of 4 to find a combination that produces -11x when multiplied. You can either use a combination of 4 and 1, or 2 and 2, since both of those numbers multiply to get 4. Just remember that one of the terms should be negative, since the term is -4.  X Research source • 3x = -1 ..... by subtracting
• 3x/3 = -1/3 ..... by dividing
• x = -1/3 ..... simplified
• x = 4 ..... by subtracting
• x = (-1/3, 4) ..... by making a set of possible, separate solutions, meaning x = -1/3, or x = 4 seem good. • So, both solutions do "check" separately, and both are verified as working and correct for two different solutions. • 4x 2 - 5x - 13 = x 2 -5
• 4x 2 - x 2 - 5x - 13 +5 = 0
• 3x 2 - 5x - 8 = 0 • {-b +/-√ (b 2 - 4ac)}/2
• {-(-5) +/-√ ((-5) 2 - 4(3)(-8))}/2(3) =
• {-(-5) +/-√ ((-5) 2 - (-96))}/2(3) • {-(-5) +/-√ ((-5) 2 - (-96))}/2(3) =
• {5 +/-√(25 + 96)}/6
• {5 +/-√(121)}/6 • (5 + 11)/6 = 16/6
• (5-11)/6 = -6/6 • x = (-1, 8/3)

## Completing the Square • 2x 2 - 9 = 12x =
• In this equation, the a term is 2, the b term is -12, and the c term is -9. • 2x 2 - 12x - 9 = 0
• 2x 2 - 12x = 9 • 2x 2 /2 - 12x/2 = 9/2 =
• x 2 - 6x = 9/2 • -6/2 = -3 =
• (-3) 2 = 9 =
• x 2 - 6x + 9 = 9/2 + 9 • x = 3 + 3(√6)/2
• x = 3 - 3(√6)/2) ## Expert Q&A • If the number under the square root is not a perfect square, then the last few steps run a little differently. Here is an example:  X Research source Thanks Helpful 1 Not Helpful 0
• As you can see, the radical sign did not disappear completely. Therefore, the terms in the numerator cannot be combined (because they are not like terms). There is no purpose, then, to splitting up the plus-or-minus. Instead, we divide out any common factors --- but ONLY if the factor is common to both of the constants AND the radical's coefficient. Thanks Helpful 1 Not Helpful 0
• If the "b" is an even number, the formula is : {-(b/2) +/- √(b/2)-ac}/a. Thanks Helpful 2 Not Helpful 0 ## You Might Also Like • ↑ https://www.mathportal.org/algebra/solving-system-of-linear-equations/elimination-method.php
• ↑ http://www.mathsisfun.com/algebra/completing-square.html To solve quadratic equations, start by combining all of the like terms and moving them to one side of the equation. Then, factor the expression, and set each set of parentheses equal to 0 as separate equations. Finally, solve each equation separately to find the 2 possible values for x. To learn how to solve quadratic equations using the quadratic formula, scroll down! Did this summary help you? Yes No

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A quadratic equation is an equation that could be written as

ax 2 + bx + c = 0

There are three basic methods for solving quadratic equations: factoring, using the quadratic formula, and completing the square.

To solve a quadratic equation by factoring,

• Put all terms on one side of the equal sign, leaving zero on the other side.
• Set each factor equal to zero.
• Solve each of these equations.

Solve x 2 – 6 x = 16.

Following the steps,

x 2 – 6 x = 16 becomes x 2 – 6 x – 16 = 0

( x – 8)( x + 2) = 0 Both values, 8 and –2, are solutions to the original equation.

Solve y 2 = – 6 y – 5.

Setting all terms equal to zero,

y 2 + 6 y + 5 = 0

( y + 5)( y + 1) = 0 To check, y 2 = –6 y – 5 A quadratic with a term missing is called an incomplete quadratic (as long as the ax 2 term isn't missing).

Solve x 2 – 16 = 0. To check, x 2 – 16 = 0 Solve x 2 + 6 x = 0. To check, x 2 + 6 x = 0 Solve 2 x 2 + 2 x – 1 = x 2 + 6 x – 5.

First, simplify by putting all terms on one side and combining like terms. Now, factor. To check, 2 x 2 + 2 x – 1 = x 2 + 6 x – 5  a, b, and c are taken from the quadratic equation written in its general form of

where a is the numeral that goes in front of x 2 , b is the numeral that goes in front of x , and c is the numeral with no variable next to it (a.k.a., “the constant”).

When using the quadratic formula, you should be aware of three possibilities. These three possibilities are distinguished by a part of the formula called the discriminant. The discriminant is the value under the radical sign, b 2 – 4 ac . A quadratic equation with real numbers as coefficients can have the following:

• Two different real roots if the discriminant b 2 – 4 ac is a positive number.
• One real root if the discriminant b 2 – 4 ac is equal to 0.
• No real root if the discriminant b 2 – 4 ac is a negative number.

Solve for x : x 2 – 5 x = –6.

Setting all terms equal to 0,

x 2 – 5 x + 6 = 0

Then substitute 1 (which is understood to be in front of the x 2 ), –5, and 6 for a , b , and c, respectively, in the quadratic formula and simplify. Because the discriminant b 2 – 4 ac is positive, you get two different real roots.

Example produces rational roots. In Example , the quadratic formula is used to solve an equation whose roots are not rational.

Solve for y : y 2 = –2y + 2.

y 2 + 2 y – 2 = 0

Then substitute 1, 2, and –2 for a , b , and c, respectively, in the quadratic formula and simplify. Note that the two roots are irrational.

Solve for x : x 2 + 2 x + 1 = 0. Since the discriminant b 2 – 4 ac is 0, the equation has one root.

The quadratic formula can also be used to solve quadratic equations whose roots are imaginary numbers, that is, they have no solution in the real number system.

Solve for x : x ( x + 2) + 2 = 0, or x 2 + 2 x + 2 = 0. Since the discriminant b 2 – 4 ac is negative, this equation has no solution in the real number system. Completing the square

A third method of solving quadratic equations that works with both real and imaginary roots is called completing the square.

• Put the equation into the form ax 2 + bx = – c . • Find the square root of both sides of the equation.
• Solve the resulting equation.

Solve for x : x 2 – 6 x + 5 = 0.

Arrange in the form of Take the square root of both sides.

x – 3 = ±2 Solve for y : y 2 + 2 y – 4 = 0. Solve for x : 2 x 2 + 3 x + 2 = 0. There is no solution in the real number system. It may interest you to know that the completing the square process for solving quadratic equations was used on the equation ax 2 + bx + c = 0 to derive the quadratic formula.

Previous Quiz: Operations with Square Roots

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## Step-By-Step Example

Example (click to try), choose your method, solve by factoring.

Example: 3x^2-2x-1=0

## Complete The Square

Example: 3x^2-2x-1=0 (After you click the example, change the Method to 'Solve By Completing the Square'.)

## Take the Square Root

Example: 2x^2=18

Example: 4x^2-2x-1=0

Need more problem types? Try MathPapa Algebra Calculator

We can help you solve an equation of the form " ax 2 + bx + c = 0 " Just enter the values of a, b and c below :

Only if it can be put in the form ax 2 + bx + c = 0 , and a is not zero .

The name comes from "quad" meaning square, as the variable is squared (in other words x 2 ).

These are all quadratic equations in disguise:

## How Does this Work?

The solution(s) to a quadratic equation can be calculated using the Quadratic Formula :

The "±" means we need to do a plus AND a minus, so there are normally TWO solutions !

The blue part ( b 2 - 4ac ) is called the "discriminant", because it can "discriminate" between the possible types of answer:

• when it is positive, we get two real solutions,
• when it is zero we get just ONE solution,
• when it is negative we get complex solutions. Quadratic equation questions are provided here for Class 10 students. A quadratic equation is a second-degree polynomial which is represented as ax 2 + bx + c = 0, where a is not equal to 0. Here, a, b and c are constants, also called coefficients and x is an unknown variable. Also, learn Quadratic Formula here.

Solving the problems based on quadratics will help students to understand the concept very well and also to score good marks in this section. All the questions are solved here step by step with a detailed explanation. In this article, we will give the definition and important formula for solving problems based on quadratic equations. The questions given here is in reference to the CBSE syllabus and NCERT curriculum.

Usually, the quadratic equation is represented in the form of ax 2 +bx+c=0, where x is the variable and a,b,c are the real numbers & a ≠ 0. Here, a and b are the coefficients of x 2 and x, respectively. So, basically, a quadratic equation is a polynomial whose highest degree is 2. Let us see some examples:

• 3x 2 +x+1, where a=3, b=1, c=1
• 9x 2 -11x+5, where a=9, b=-11, c=5

If we solve any quadratic equation, then the value we obtain are called the roots of the equation. Since the degree of the quadratic equation is two, therefore we get here two solutions and hence two roots.

There are different methods to find the roots of quadratic equation, such as:

• Factorisation
• Completing the square

The quadratic formula to find the roots of the quadratic equation is given by:

$$\begin{array}{l}x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\end{array}$$

Where b 2 -4ac is called the discriminant of the equation.

Based on the discriminant value, there are three possible conditions, which defines the nature of roots as follows:

• two distinct real roots, if b 2 – 4ac > 0
• two equal real roots, if b 2 – 4ac = 0
• no real roots, if b 2 – 4ac < 0

Also, learn quadratic equations for class 10 here.

## Quadratic Equations Problems and Solutions

1. Rahul and Rohan have 45 marbles together. After losing 5 marbles each, the product of the number of marbles they both have now is 124. How to find out how many marbles they had to start with.

Solution: Say, the number of marbles Rahul had be x.

Then the number of marbles Rohan had = 45 – x.

The number of marbles left with Rahul after losing 5 marbles = x – 5

The number of marbles left with Rohan after losing 5 marbles = 45 – x – 5 = 40 – x

The product of number of marbles = 124

(x – 5) (40 – x) = 124

40x – x 2 – 200 + 5x = 124

– x 2 + 45x – 200 = 124

x 2 – 45x + 324 = 0

This represents the quadratic equation. Hence by solving the given equation for x, we get;

x = 36 and x = 9

So, the number of marbles Rahul had is 36 and Rohan had is 9 or vice versa.

2. Check if x(x + 1) + 8 = (x + 2) (x – 2) is in the form of quadratic equation.

Solution: Given,

x(x + 1) + 8 = (x + 2) (x – 2)

Cancel x 2 both the sides.

Since, this expression is not in the form of ax 2 +bx+c, hence it is not a quadratic equation.

3. Find the roots of the equation 2x 2 – 5x + 3 = 0 using factorisation.

2x 2 – 5x + 3 = 0

2x 2 – 2x-3x+3 = 0

2x(x-1)-3(x-1) = 0

(2x-3) (x-1) = 0

2x-3 = 0; x = 3/2

(x-1) = 0; x=1

Therefore, 3/2 and 1 are the roots of the given equation.

4. Solve the quadratic equation 2x 2 + x – 300 = 0 using factorisation.

Solution: 2x 2 + x – 300 = 0

2x 2 – 24x + 25x – 300 = 0

2x (x – 12) + 25 (x – 12) = 0

(x – 12)(2x + 25) = 0

x-12=0; x=12

(2x+25) = 0; x=-25/2 = -12.5

Therefore, 12 and -12.5 are two roots of the given equation.

5. Solve the equation x 2 +4x-5=0.

x 2 + 4x – 5 = 0

x 2 -1x+5x-5 = 0

x(x-1)+5(x-1) =0

(x-1)(x+5) =0

Hence, (x-1) =0, and (x+5) =0

similarly, x+5 = 0

x=-5 & x=1

6. Solve the quadratic equation 2x 2 + x – 528 = 0, using quadratic formula.

Solution: If we compare it with standard equation, ax 2 +bx+c = 0

a=2, b=1 and c=-528

Hence, by using the quadratic formula:

Now putting the values of a,b and c.

x=64/4 or x=-66/4

x=16 or x=-33/2

7. Find the roots of x 2 + 4x + 5 = 0, if any exist, using quadratic formula.

Solution: To check whether there are real roots available for the quadratic equation, we need the find the discriminant value.

D = b 2 -4ac = 4 2 – 4(1)(5) = 16-20 = -4

Since the square root of -4 will not give a real number. Hence there is no real roots for the given equation.

8. Find the discriminant of the equation: 3x 2 -2x+⅓ = 0.

Solution: Here, a = 3, b=-2 and c=⅓

Hence, discriminant, D = b 2 – 4ac

D = (-2) 2 -4(3)(⅓)

## Video Lesson ## Practice Questions

Solve these quadratic equations and find the roots.

• x 2 -5x-14=0         [Answer: x=-2 & x=7]
• X 2  = 11x -28       [Answer: x=4 & x = 7]
• 6x 2 – x = 5            [Answer: x=-⅚ & x = 1]
• 12x 2 = 25x          [Answer: x=0 & x=25/12]

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Number line.

• ax^2+bx+c=0
• x^2+2x+1=3x-10
• 2x^2+4x-6=0
• How do you calculate a quadratic equation?
• To solve a quadratic equation, use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a).
• What is the quadratic formula?
• The quadratic formula gives solutions to the quadratic equation ax^2+bx+c=0 and is written in the form of x = (-b ± √(b^2 - 4ac)) / (2a)
• Does any quadratic equation have two solutions?
• There can be 0, 1 or 2 solutions to a quadratic equation. If the discriminant is positive there are two solutions, if negative there is no solution, if equlas 0 there is 1 solution.
• What is quadratic equation in math?
• In math, a quadratic equation is a second-order polynomial equation in a single variable. It is written in the form: ax^2 + bx + c = 0 where x is the variable, and a, b, and c are constants, a ≠ 0.
• How do you know if a quadratic equation has two solutions?
• A quadratic equation has two solutions if the discriminant b^2 - 4ac is positive.

• High School Math Solutions – Quadratic Equations Calculator, Part 1 A quadratic equation is a second degree polynomial having the general form ax^2 + bx + c = 0, where a, b, and c... Read More
• First-order ODE
• Introduction Differential equations Order Linearity
• Direction field method Implicit/explicit form Method
• Separation of variables Separable Reduction to separable form
• Equilibrium Characterization Stability
• Linear 1st order ODE technique Standard form General solution Reduction to linear form
• ODE existence and uniqueness Existence theorem Uniqueness theorem
• Initial value numerical methods Error Taylor series Stability Euler methods Runge-Kutta methods
• System of linear ODEs Definition Characteristic equation Eigenvector/eigenvalue System of homogeneous ODEs

## First-order Ordinary Differential Equations cheatsheet Star

By Afshine Amidi and Shervine Amidi

## Introduction

Differential Equations A differential equation is an equation containing derivatives of a dependent variable $y$ with respect to independent variables $x$. In particular,

- Ordinary Differential Equations ( ODE ) are differential equations having one independent variable.

- Partial Differential Equations ( PDE ) are differential equations having two or more independent variables.

Order An ODE is said to be of order $n$ if the highest derivative of the unknown function in the equation is the $n^{th}$ derivative with respect to the independent variable.

Linearity An ODE is said to be linear only if the function $y$ and all of its derivatives appear by themselves. Thus, it is of the form:

## Direction Field Method

Implicit form The implicit form of an ODE is where $y'$ is not separated from the remaining terms of the ODE. It is of the form:

Remark: Sometimes, $y'$ cannot be separated from the other terms and the implicit form is the only one that we can write.

Explicit form The explicit form of an ODE is where $y'$ is separated from the remaining terms of the ODE. It is of the form:

Direction field method The direction field method is a graphical representation for the solution of ODE $y'=f(x,y)$ without actually solving for $y(x)$. Here is the procedure:

## Separation of Variables

Separable An ODE is said to be separable if it can be written in the form:

Reduction to separable form The following table sums up the variable changes that allow us to change the ODE $y'=f(x,y)$ to $u'=g(x,u)$ that is separable.

## Equilibrium

Characterization In order for an ODE to have equilibrium solutions, it must be (1) autonomous and (2) have a value $y^*$ that makes the derivative equal to 0, i.e:

Stability Equilibrium solutions can be classified into 3 categories:

- Unstable : solutions run away with any small change to the initial conditions. - Stable : any small perturbation leads the solutions back to that solution. - Semi-stable : a small perturbation is stable on one side and unstable on the other.

## Linear first-order ODE technique

Standard form The standard form of a first-order linear ODE is expressed with $p(x), r(x)$ known functions of $x$, such that:

Remark: If $r=0$, then the ODE is homogenous, and if $r\neq0$, then the ODE is inhomogeneous.

General solution The general solution $y$ of the standard form can be decomposed into a homogenous part $y_h$ and a particular part $y_p$ and is expressed in terms of $p(x), r(x)$ such that:

Remark 1: Here, for any function $p$, the notation $\displaystyle\int pdx$ denotes the primitive of $p$ without additive constant.

Remark 2: The term $e^{-\int pdx}$ is called the basis of the ODE and $e^{\int pdx}$ is called the integrating factor.

Reduction to linear form The one-line table below sums up the change of variables that we apply in order to have a linear form:

## Existence and uniqueness of an ODE

Here, we are given an ODE $y'=f(x,y)$ with initial conditions $y(x_0)=y_0$.

Existence theorem If $f(x,y)$ is continuous at all points in a rectangular region containing $(x_0,y_0)$, then $y'=f(x,y)$ has at least one solution $y(x)$ passing through $(x_0,y_0)$.

Remark: If the condition does not apply, then we cannot say anything about existence.

Uniqueness theorem If both $f(x,y)$ and $\frac{\partial f}{\partial y}(x,y)$ are continuous at all points in a rectangular region containing $(x_0,y_0)$, then $y'=f(x,y)$ has a unique solution $y(x)$ passing through $(x_0,y_0)$.

Remark: If the condition does not apply, then we cannot say anything about uniqueness.

## Numerical methods for ODE - Initial value problems

In this section, we would like to find $y(t)$ for the interval $[0,t_f]$ that we divide into $N+1$ equally-spaced points $t_0< t_1 < ... < t_N = t_f$, such that:

Error In order to assess the accuracy of a numerical method, we define its local and global errors $\epsilon_{\textrm{local}}, \epsilon_{\textrm{global}}$ as follows:

Remark 1: If $\epsilon_{\textrm{local}}=O(h^k)$, then $\epsilon_{\textrm{global}}=O(h^{k-1})$.

Remark 2: When we talk about the 'error' of a method, we refer to its global error.

Taylor series The Taylor series giving the exact expression of $y_{n+1}$ in terms of $y_n$ and its derivatives is:

We can also have an expression of $y_n$ in terms of $y_{n+1}$ and its derivatives:

Stability The stability analysis of any ODE solver algorithm is performed on the model problem , defined by:

which gives $y_n=y_0\sigma^n$, for which $h$ verifies the condition $|\sigma(h)| < 1$.

Euler methods The Euler methods are numerical methods that aim at estimating the solution of an ODE:

Runge-Kutta methods The table below sums up the most commonly used Runge-Kutta methods:

## System of Linear ODEs

Definition A system of $n$ first order linear ODEs

can be written in matrix form as:

where $A=\left(\begin{array}{ccc}a_{11}& \cdots& a_{1n}\\\vdots& \ddots & \vdots\\a_{n1}& \cdots& a_{nn}\end{array}\right)$ and $\vec{y}=\left(\begin{array}{c}y_1\\\vdots\\y_n\end{array}\right)$

Characteristic equation The characteristic equation of a linear system of $n$ equations represented by $A$ is given by:

For $n=2$, this equation can be written as:

Eigenvector, eigenvalue The roots $\lambda$ of the characteristic equation are the eigenvalues of $A$. The solutions $\vec{v}$ of the equation $A\vec{v}=\lambda I$ are called the eigenvectors associated with the eigenvalue $\lambda$.

System of homogeneous ODEs The resolution of the system of 2 homogeneous linear ODEs $\vec{y}'=A\vec{y}$ is detailed in the following table:

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## Computer Science > Computational Complexity

Title: the complexity of computing kkt solutions of quadratic programs.

Abstract: It is well known that solving a (non-convex) quadratic program is NP-hard. We show that the problem remains hard even if we are only looking for a Karush-Kuhn-Tucker (KKT) point, instead of a global optimum. Namely, we prove that computing a KKT point of a quadratic polynomial over the domain $[0,1]^n$ is complete for the class CLS = PPAD$\cap$PLS.

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Both individuals and organizations that work with arXivLabs have embraced and accepted our values of openness, community, excellence, and user data privacy. arXiv is committed to these values and only works with partners that adhere to them. #### IMAGES  3. Solving A Quadratic Equation By Factoring 4. Quadratic Equation Worksheet /Problem with Solution 5. Math 10: CHAPTER 3 : SOLUTION OF QUADRATIC EQUATIONS 6. How to Write the Quadratic Equation with the Given Solution Set #### VIDEO

2. How To Solve Quadratic Equations By Factoring

3. Solving a quadratic equation by factoring

4. Writing Quadratic Equations In Standard Form Given The Solution

5. Introduction to the quadratic equation

6. Completing The Square Method and Solving Quadratic Equations

Problem 1: Solve the quadratic equation using the quadratic formula. ${x^2}\, - \,8x + 12 = 0$ Answer

2. Algebra

Here is a set of practice problems to accompany the Quadratic Equations - Part I section of the Solving Equations and Inequalities chapter of the notes for Paul Dawkins Algebra course at Lamar University.

Number of solutions of quadratic equations Quadratic formula review Discriminant review Math > Algebra 1 > Quadratic functions & equations > The quadratic formula Quadratic formula Google Classroom Solve. 6 + 2 x 2 − 3 x = 8 x 2 Choose 1 answer: x = 3, − 1 2 A x = 3, − 1 2 x = 5 ± 57 16 B x = 5 ± 57 16 x = 1 ± 17 − 4 C x = 1 ± 17 − 4 x = − 4 ± 34 3

4. 9.6: Solve Applications of Quadratic Equations

This is a quadratic equation, rewrite it in standard form. Solve the equation using the Quadratic Formula. Identify the values of $$a, b, c$$. Write the Quadratic Formula. Then substitute in the values of $$a,b,c$$. Simplify. Figure 9.5.26: Rewrite to show two solutions. Approximate the answer with a calculator. Step 6: Check the answer. The ...

5. Solving quadratic equations by factoring (article)

You may have also solved some quadratic equations, which include the variable raised to the second power, by taking the square root from both sides. In this lesson, you will learn a new way to solve quadratic equations. Specifically you will learn how to solve factored equations like ( x − 1) ( x + 3) = 0 and

This topic covers: - Solving quadratic equations - Graphing quadratic functions - Features of quadratic functions - Quadratic equations/functions word problems - Systems of quadratic equations - Quadratic inequalities Parabolas intro Learn Parabolas intro Practice Parabolas intro 4 questions Practice Quadratic factored form Learn

Solve by completing the square: Non-integer solutions. Worked example: completing the square (leading coefficient ≠ 1) Solving quadratics by completing the square: no solution. Proof of the quadratic formula. Solving quadratics by completing the square. Completing the square review. Quadratic formula proof review.

8. 4 Ways to Solve Quadratic Equations

A quadratic equation is a polynomial equation in a single variable where the highest exponent of the variable is 2.  There are three main ways to solve quadratic equations: 1) to factor the quadratic equation if you can do so, 2) to use the quadratic formula, or 3) to complete the square.

The quadratic formula helps us solve any quadratic equation. First, we bring the equation to the form ax²+bx+c=0, where a, b, and c are coefficients. Then, we plug these coefficients in the formula: (-b±√ (b²-4ac))/ (2a) . See examples of using the formula to solve a variety of equations. Created by Sal Khan.

How To Solve Them? The " solutions " to the Quadratic Equation are where it is equal to zero. They are also called " roots ", or sometimes " zeros " There are usually 2 solutions (as shown in this graph). And there are a few different ways to find the solutions: We can Factor the Quadratic (find what to multiply to make the Quadratic Equation)

Worked example. First we need to identify the values for a, b, and c (the coefficients). First step, make sure the equation is in the format from above, a x 2 + b x + c = 0 : is what makes it a quadratic). . . Therefore x = 3 or x = − 7 .

12. Solving quadratics by completing the square

To factor the equation, you need to first follow this equation: x^ 2 + 2ax + a^2. In x^2 +5x = 3/4, The a^2 is missing. To figure out the a, you need to take the 5 and divide it by 2 (because 2ax), which becomes 5/2. a=5/2. Then you need to square it, (because a^2) which becomes 5^2/2^2. 5x5 is 25, and 2x2 is 4, so the a^2 is 25/4.

Substituting in the quadratic formula, Since the discriminant b 2 - 4 ac is 0, the equation has one root. The quadratic formula can also be used to solve quadratic equations whose roots are imaginary numbers, that is, they have no solution in the real number system. Example 9. Solve for x: x( x + 2) + 2 = 0, or x 2 + 2 x + 2 = 0.

Plots of quadratic function y = ax2 + bx + c, varying each coefficient separately while the other coefficients are fixed (at values a = 1, b = 0, c = 0) A quadratic equation with real or complex coefficients has two solutions, called roots. These two solutions may or may not be distinct, and they may or may not be real. Factoring by inspection

There are different methods you can use to solve quadratic equations, depending on your particular problem. Solve By Factoring. Example: 3x^2-2x-1=0. Complete The Square. Example: 3x^2-2x-1=0 (After you click the example, change the Method to 'Solve By Completing the Square'.) Take the Square Root. Example: 2x^2=18. Quadratic Formula

16. Quadratic Equations: Problems with Solutions

Quadratic Equations: Problems with Solutions Problem 1 How many real roots does the equation have? \displaystyle x^2 + 3x + 4 = 0 x2 +3x+4 = 0 Problem 2 What is the value of the greater root of the equation \displaystyle x^2-5x+4=0 x2 −5x+4 = 0 ? Problem 3 What is the value of the lesser root of the equation \displaystyle x^2-3x+2=0 x2 −3x+2 = 0 ?

17. Real World Examples of Quadratic Equations

Let us solve it ... There are many ways to solve it, here we will factor it using the "Find two numbers that multiply to give a×c, and add to give b " method in Factoring Quadratics: a×c = −15, and b = −14. The factors of −15 are: −15, −5, −3, −1, 1, 3, 5, 15 By trying a few combinations we find that −15 and 1 work (−15×1 = −15, and −15+1 = −14)

The solution (s) to a quadratic equation can be calculated using the Quadratic Formula: The "±" means we need to do a plus AND a minus, so there are normally TWO solutions ! The blue part ( b2 - 4ac) is called the "discriminant", because it can "discriminate" between the possible types of answer: when it is negative we get complex solutions.

Quadratic Equations Problems and Solutions 1. Rahul and Rohan have 45 marbles together. After losing 5 marbles each, the product of the number of marbles they both have now is 124. How to find out how many marbles they had to start with. Solution: Say, the number of marbles Rahul had be x.

20. 9.6: Introduction to Complex Numbers and Complex Solutions

Solution: Step 1: Write the quadratic equation in standard form. Step 2: Identify a, b, and c for use in the quadratic formula. Here. Step 3: Substitute the appropriate values into the quadratic formula and then simplify. Answer: The solution is $$\frac{3}{2} \pm \frac{1}{2} i$$. The check is optional.

The quadratic formula gives solutions to the quadratic equation ax^2+bx+c=0 and is written in the form of x = (-b ± √ (b^2 - 4ac)) / (2a) Does any quadratic equation have two solutions? There can be 0, 1 or 2 solutions to a quadratic equation.

22. Module 4 (M4)

A formula can be used to solve any equation in the form $$\mathbf{ax^2 + bx + c = 0}$$. If the quadratic expression can be factorised, it is generally quicker to use the factors method.

Problems Involving the Quadratic Formula. First comes the quadratic equation, then comes the quadratic formula. The quadratic formula is the solution to the quadratic equation: $ax^2+bx+c=0$ in which. x is the variable whose value is sought, and a, b, and c are constants. The goal is to find the value of x that makes the left side 0.

24. CME 102

Stability Equilibrium solutions can be classified into 3 categories: - Unstable: solutions run away with any small change to the initial conditions. - Stable: any small perturbation leads the solutions back to that solution. - Semi-stable: a small perturbation is stable on one side and unstable on the other. Linear first-order ODE technique. Standard form The standard form of a first-order ...

25. The Complexity of Computing KKT Solutions of Quadratic Programs

The Complexity of Computing KKT Solutions of Quadratic Programs. John Fearnley, Paul W. Goldberg, Alexandros Hollender, Rahul Savani. It is well known that solving a (non-convex) quadratic program is NP-hard. We show that the problem remains hard even if we are only looking for a Karush-Kuhn-Tucker (KKT) point, instead of a global optimum.

26. Algebra

Section 2.6 : Quadratic Equations - Part II. For problems 1 - 3 complete the square. x2 +8x x 2 + 8 x Solution. u2 −11u u 2 − 11 u Solution. 2z2 −12z 2 z 2 − 12 z Solution. For problems 4 - 8 solve the quadratic equation by completing the square. t2−10t+34 = 0 t 2 − 10 t + 34 = 0 Solution. v2 +8v−9 = 0 v 2 + 8 v − 9 = 0 ...