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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

A quadratic equation is an equation with a
as its highest power term. For example, in the quadratic equation $3{x}^{2}-5x-2=0$:
• $x$ is the variable, which represents a number whose value we don't know yet.
• The ${}^{2}$ is the power or exponent. An exponent of $2$ means the variable is
.
• $3$ and $-5$ are the coefficients, or constant multiples of ${x}^{2}$ and $x$. $3{x}^{2}$ is a single
, as is $-5x$.
• $-2$ is a constant term.
In this lesson, we'll learn to:
1. Solve quadratic equations in several different ways
2. 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

Solving quadratics by taking square rootsSee video transcript

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

Quadratic equations without $x$-terms such as $2{x}^{2}=32$ can be solved without setting a quadratic expression equal to $0$. Instead, we can isolate ${x}^{2}$ and use the square root operation to solve for $x$.
When 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:
• ${2}^{2}\stackrel{✓}{=}4$
• $\left(-2{\right)}^{2}\stackrel{✓}{=}4$
When solving quadratic equations without $x$-terms:
1. Isolate ${x}^{2}$.
2. Take the square root of both sides of the equation. Both the positive and negative square roots are solutions.

Example: What values of $x$ satisfy the equation $2{x}^{2}=18$ ?

### Try it!

TRY: identify the steps to solving a quadratic equation
${x}^{2}-3=13$
We can solve the quadratic equation above by first
both sides of the equation, which gives us the equation ${x}^{2}=16$.
Next, we can take the square root of both sides of the equation, which gives us the solution(s)
.

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

### Zero product property

Zero product propertySee video transcript

### Zero product property and factored quadratic equations

The zero product property states that if $ab=0$, then either $a$ or $b$ is equal to $0$.
The zero product property lets us solve factored quadratic equations by solving two linear equations. For a quadratic equation such as $\left(x-5\right)\left(x+2\right)=0$, 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:
1. Set each factor equal to $0$.
2. 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 $\left(x-4\right)\left(3x+1\right)=0$ ?

### Try it!

TRY: use factors to determine the solutions
$2{x}^{2}+x-3=\left(x-1\right)\left(2x+3\right)$
The equation above shows the factors of the quadratic expression $2{x}^{2}+x-3$. Which of the following equations, when solved, give us the solutions to the equation $2{x}^{2}+x-3=0$ ?

## How do I solve quadratic equations by factoring?

Solving quadratics by factoringSee video transcript

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 $\left(x+a\right)\left(x+b\right)$, where $a$ and $b$ fulfill the following criteria:
• The sum of $a$ and $b$ is equal to the coefficient of the $x$-term in the unfactored quadratic expression.
• The product of $a$ and $b$ is equal to the constant term of the unfactored quadratic expression.
For example, we can solve the equation ${x}^{2}-2x-3=0$ by factoring ${x}^{2}-2x-3$ into $\left(x+a\right)\left(x+b\right)$, where:
• $a+b$ is equal to the coefficient of the $x$-term, $-2$.
• $ab$ is equal to the constant term, $-3$.
$-3$ and $1$ would work:
• $-3+1=-2$
• $\left(-3\right)\left(1\right)=-3$
This means we can rewrite ${x}^{2}-2x-3=0$ as $\left(x-3\right)\left(x+1\right)=0$ 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 ${x}^{2}+bx+c=0$:
1. Rewrite the quadratic expression as the product of two factors. The two factors are linear expressions with an $x$-term and a constant term. The sum of the constant terms is equal to $b$, and the product of the constant terms is equal to $c$.
2. Set each factor equal to $0$.
3. 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 ${x}^{2}+4x-5=0$ ?

### Try it!

Try: match the equivalent quadratic expressions
Match each factored expression to its equivalent unfactored expression in the table below.

## How do I use 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 $a{x}^{2}+bx+c=0$:
$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
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?

1. Rewrite the equation in the form $a{x}^{2}+bx+c=0$.
2. Substitute the values of $a$, $b$, and $c$ into the quadratic formula, shown below.
$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
1. Evaluate $x$.

Example: What are the solutions to the equation ${x}^{2}-6x=9$ ?

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

### Try it!

Try: set up for the quadratic formula
$7{x}^{2}+6x-1=0$
If we want to use the quadratic formula to solve the equation above, what are the values of $a$, $b$, and $c$ we should plug into the quadratic formula?
$a=$
$b=$
$c=$

Try: use the discriminant to find the number of solutions
$7{x}^{2}+6x-1=0$
The value of the
for the quadratic equation above is
.
Because the discriminant is
, the equation has
.

Try: substitute into the quadratic formula
$7{x}^{2}+6x-1=0$
Which of the following expressions, when evaluated, gives the solutions to the equation above?

Practice: Solve quadratic equation using square root
If $\frac{1}{2}{x}^{2}=32$ and $x>0$, what is the value of $x$ ?

Practice: Solve quadratic equation by factoring
${x}^{2}+x-56=0$
What are the solutions to the equation above?

Which of the following values of $x$ satisfy the equation $-3{x}^{2}+12x+4=0$ ?

Practice: Determine the condition for one real solution
If $a{x}^{2}+8x+1=0$, for what value of $a$ does the equation have exactly one real solution?

## Things to remember

For $a{x}^{2}+bx+c=0$:
$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
• If ${b}^{2}-4ac>0$, then the equation has $2$ unique real solutions.
• If ${b}^{2}-4ac=0$, then the equation has $1$ unique real solution.
• If ${b}^{2}-4ac<0$, then the equation has no real solution.

## Want to join the conversation?

• why try to factor if i can just use a formula
• 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.
• I love his calculator
• can't I just use the calculator to see the solutions for the quadratic equation?
• loved it

you can learn anything. let's do this!
• 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
• The question asks for exactly one real solution so you should equate the discriminate value to zero
• how do u know when to use the quadratic formula and when to solve by factoring?
• 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?