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### Course: Digital SAT Math > Unit 8

Lesson 7: Solving quadratic equations: medium# Solving quadratic equations | Lesson

A guide to solving quadratic equations on the digital SAT

## What are quadratic equations?

A $3{x}^{2}-5x-2=0$ :

**quadratic equation**is an equation with a as its highest power term. For example, in the quadratic equation is the$x$ **variable**, which represents a number whose value we don't know yet.- The
is the${}^{2}$ **power**or**exponent**. An exponent of means the variable is .$2$ and$3$ are the$-5$ **coefficients**, or constant multiples of and${x}^{2}$ .$x$ is a single , as is$3{x}^{2}$ .$-5x$ is a$-2$ **constant**term.

In this lesson, we'll learn to:

- Solve quadratic equations in several different ways
- Determine the number of solutions to a quadratic equation without solving

**You can learn anything. Let's do this!**

## How do I solve quadratic equations using square roots?

### Solving quadratics by taking square roots

### When can I solve by taking square roots?

Quadratic equations without $x$ -terms such as $2{x}^{2}=32$ can be solved $0$ . Instead, we can isolate ${x}^{2}$ and use the square root operation to solve for $x$ .

*without*setting a quadratic expression equal toWhen solving quadratic equations by taking square roots,

**both the positive and negative square roots are solutions to the equation**. This is because when we*square*a solution, the result is*always positive*.For example, for the equation ${x}^{2}=4$ , both $2$ and $-2$ are solutions:

${2}^{2}\check{=}4$ $(-2{)}^{2}\check{=}4$

When solving quadratic equations without $x$ -terms:

- Isolate
.${x}^{2}$ - Take the square root of both sides of the equation. Both the positive and negative square roots are solutions.

**Example:**What values of

### Try it!

## What is the zero product property, and how do I use it to solve quadratic equations?

### Zero product property

### Zero product property and factored quadratic equations

The $ab=0$ , then either $a$ or $b$ is equal to $0$ .

**zero product property**states that ifThe zero product property lets us solve factored quadratic equations by solving two linear equations. For a quadratic equation such as $(x-5)(x+2)=0$ , we know that either $x-5=0$ or $x+2=0$ . Solving these two linear equations gives us the two solutions to the quadratic equation.

To solve a factored quadratic equation using the zero product property:

- Set each factor equal to
.$0$ - Solve the equations from Step 1. The solutions to the linear equations are also solutions to the quadratic equation.

**Example:**What are the solutions to the equation

### Try it!

## How do I solve quadratic equations by factoring?

### Solving quadratics by factoring

### Solving factorable quadratic equations

If we can write a quadratic expression as the product of two linear expressions (factors), then we can use those linear expressions to calculate the solutions to the quadratic equation.

In this lesson, we'll focus on factorable quadratic equations with $1$ as the coefficient of the ${x}^{2}$ term, such as ${x}^{2}-2x-3=0$ . For more advanced factoring techniques, including special factoring and factoring quadratic expressions with ${x}^{2}$ coefficients other than $1$ , check out the

**Factoring quadratic and polynomial expressions**lesson.Recognizing factors of quadratic expressions takes practice. The factors will be in the form $(x+a)(x+b)$ , where $a$ and $b$ fulfill the following criteria:

- The
*sum*of and$a$ is equal to the coefficient of the$b$ -term in the unfactored quadratic expression.$x$ - The
*product*of and$a$ is equal to the constant term of the unfactored quadratic expression.$b$

For example, we can solve the equation ${x}^{2}-2x-3=0$ by factoring ${x}^{2}-2x-3$ into $(x+a)(x+b)$ , where:

is equal to the coefficient of the$a+b$ -term,$x$ .$-2$ is equal to the constant term,$ab$ .$-3$

$-3+1=-2$ $(-3)(1)=-3$

This means we can rewrite ${x}^{2}-2x-3=0$ as $(x-3)(x+1)=0$ and solve the quadratic equation using the zero product property. Keep mind that $a$ and $b$ are

*not*themselves solutions to the quadratic equation!When solving factorable quadratic equations in the form ${x}^{2}+bx+c=0$ :

- Rewrite the quadratic expression as the product of two factors. The two factors are linear expressions with an
-term and a constant term. The sum of the constant terms is equal to$x$ , and the product of the constant terms is equal to$b$ .$c$ - Set each factor equal to
.$0$ - Solve the equations from Step 2. The solutions to the linear equations are also solutions to the quadratic equation.

**Example:**What are the solutions to the equation

### Try it!

## How do I use the quadratic formula?

### The quadratic formula

### Using the quadratic formula to solve equations and determine the number of solutions

Not all quadratic expressions are factorable, and not all factorable quadratic expressions are easy to factor.

**The quadratic formula**gives us a way to solve any quadratic equation as long as we can plug the correct values into the formula and evaluate.For ${a}{x}^{2}+{b}x+{c}=0$ :

**Note:**the quadratic formula is

*not*provided in the reference section of the SAT! You'll have to memorize the formula to use it.

### What are the steps?

To solve a quadratic equation using the quadratic formula:

- Rewrite the equation in the form
.$a{x}^{2}+bx+c=0$ - Substitute the values of
,$a$ , and$b$ into the quadratic formula, shown below.$c$

- Evaluate
.$x$

**Example:**What are the solutions to the equation

The ${b}^{2}-4ac$ portion of the quadratic formula is called the $b-4ac$ tells us the

**discriminant**. The value of**number of unique real solutions**the equation has:- If
, then${b}^{2}-4ac>0$ is a$\sqrt{{b}^{2}-4ac}$ **real number**, and the quadratic equation has**two real solutions**, and$\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}$ .$\frac{-b+\sqrt{{b}^{2}-4ac}}{2a}$ - If
, then${b}^{2}-4ac=0$ is also$\sqrt{{b}^{2}-4ac}$ , and the quadratic formula simplifies to$0$ , which means the quadratic equation has$\frac{-b}{2a}$ **one real solution**. - If
, then${b}^{2}-4ac<0$ is an$\sqrt{{b}^{2}-4ac}$ **imaginary number**, which means the quadratic equation has**no real solutions**.

### Try it!

## Your turn!

## Things to remember

For $a{x}^{2}+bx+c=0$ :

- If
, then the equation has${b}^{2}-4ac>0$ unique real solutions.$2$ - If
, then the equation has${b}^{2}-4ac=0$ unique real solution.$1$ - If
, then the equation has no real solution.${b}^{2}-4ac<0$

## Want to join the conversation?

- why try to factor if i can just use a formula(16 votes)
- Hi!

The formula is the easiest way to solve the quadratic equation. It works for all types of equations and can even be used to find out if the equation has a real solution or not. But sometimes using it may be complicated and you may find it easier to directly factorize it. But that's totally upto you. If you're already familiar with the formula and feel more comfortable with it, go ahead and use it! Always use the most time-saving method on the SAT.(41 votes)

- I love his calculator(25 votes)
- can't I just use the calculator to see the solutions for the quadratic equation?(12 votes)
- loved it

you can learn anything. let's do this!(10 votes) - Can someone please explain to me the last quadratic equation question. When I look at khan academy explanation, it jumps from -12 to just to 2 at some point. Even when I do the question with calculator or using photomath, Khan Academy's answers just seem to be wrong. Am I missing something, please help me(7 votes)
- The question asks for exactly one real solution so you should equate the discriminate value to zero(2 votes)

- how do u know when to use the quadratic formula and when to solve by factoring?(4 votes)
- use the formula when the q asks to factor a huge middle term(1 vote)

- in the last video, the third example; what happened to 2 in 12+/- 2 root 39?(2 votes)
- he divided 2 by 6(2 votes)

- in the question with '7x^2+6x-1=0'. wouldn't the solution to the discriminant be 4? because -4(7)(-1) becomes positive 28? because -4*7= -28, -28*-1= 28. or can somebody help me with what i'm doing wrong?(2 votes)
- Hi, I'm not sure if I understood you right but the discriminant is 64 because 36 - (-28) = 64(2 votes)

- Won't 6^2 −4(7)(−1) result in 36 - (28)?(2 votes)
- in the last quadratic example breakdown when we have -12+/_ 2 sq root 39 all divided by -6, why did Sal divide a the 12,6 and 2 by 2? rather than dividing 2 into itself and then into -6 which will produce -3. -3 would then go into -12 produce 4 then continue?(2 votes)