Main content

## SAT

### Unit 6: Lesson 4

Passport to Advanced Math: lessons by skill- Solving quadratic equations | Lesson
- Interpreting nonlinear expressions | Lesson
- Quadratic and exponential word problems | Lesson
- Manipulating quadratic and exponential expressions | Lesson
- Radicals and rational exponents | Lesson
- Radical and rational equations | Lesson
- Operations with rational expressions | Lesson
- Operations with polynomials | Lesson
- Polynomial factors and graphs | Lesson
- Graphing quadratic functions | Lesson
- Graphing exponential functions | Lesson
- Linear and quadratic systems | Lesson
- Structure in expressions | Lesson
- Isolating quantities | Lesson
- Function Notation | Lesson

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# Solving quadratic equations | Lesson

## What are quadratic equations, and how frequently do they appear on the test?

A

**quadratic equation**is an equation with a as its highest power term. For example, in the quadratic equation 3, x, squared, minus, 5, x, minus, 2, equals, 0:- x is the
**variable**, which represents a number whose value we don't know yet. - The squared is the
**power**or**exponent**. An exponent of 2 means the variable is . - 3 and minus, 5 are the
**coefficients**, or constant multiples of x, squared and x. 3, x, squared is a single , as is minus, 5, x. - minus, 2 is a
**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

On your official SAT, you'll likely see

**0 to 2 questions**that test your ability to solve quadratic equations—more when you include quadratic word problems, linear and quadratic systems, and graphing quadratic functions.**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, squared, equals, 32 can be solved

*without*setting a quadratic expression equal to 0. Instead, we can isolate x, squared 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, squared, equals, 4, both 2 and minus, 2 are solutions:

- 2, squared, equals, start superscript, \checkmark, end superscript, 4
- left parenthesis, minus, 2, right parenthesis, squared, equals, start superscript, \checkmark, end superscript, 4

When solving quadratic equations without x-terms:

- Isolate x, squared.
- 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, squared, equals, 18 ?

### 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

**zero product property**states that if a, b, equals, 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 parenthesis, x, minus, 5, right parenthesis, left parenthesis, x, plus, 2, right parenthesis, equals, 0, we know that either x, minus, 5, equals, 0 or x, plus, 2, equals, 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 left parenthesis, x, minus, 4, right parenthesis, left parenthesis, 3, x, plus, 1, right parenthesis, equals, 0 ?

### 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, squared term, such as x, squared, minus, 2, x, minus, 3, equals, 0. For more advanced factoring techniques, including special factoring and factoring quadratic expressions with x, squared coefficients other than 1, check out the

**Structure in expressions**lesson.Recognizing factors of quadratic expressions takes practice. The factors will be in the form left parenthesis, x, plus, a, right parenthesis, left parenthesis, x, plus, b, right parenthesis, 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, squared, minus, 2, x, minus, 3, equals, 0 by factoring x, squared, minus, 2, x, minus, 3 into left parenthesis, x, plus, a, right parenthesis, left parenthesis, x, plus, b, right parenthesis, where:

- a, plus, b is equal to the coefficient of the x-term, minus, 2.
- a, b is equal to the constant term, minus, 3.

minus, 3 and 1 would work:

- minus, 3, plus, 1, equals, minus, 2
- left parenthesis, minus, 3, right parenthesis, left parenthesis, 1, right parenthesis, equals, minus, 3

This means we can rewrite x, squared, minus, 2, x, minus, 3, equals, 0 as left parenthesis, x, minus, 3, right parenthesis, left parenthesis, x, plus, 1, right parenthesis, equals, 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, squared, plus, b, x, plus, c, equals, 0:

- 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.
- 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 x, squared, plus, 4, x, minus, 5, equals, 0 ?

### 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 start color #7854ab, a, end color #7854ab, x, squared, plus, start color #ca337c, b, end color #ca337c, x, plus, start color #208170, c, end color #208170, equals, 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, squared, plus, b, x, plus, c, equals, 0.
- Substitute the values of a, b, and c into the quadratic formula, shown below.

- Evaluate x.

**Example:**What are the solutions to the equation x, squared, minus, 6, x, equals, 9 ?

The b, squared, minus, 4, a, c portion of the quadratic formula is called the

**discriminant**. The value of b, minus, 4, a, c tells us the**number of unique real solutions**the equation has:- If b, squared, minus, 4, a, c, is greater than, 0, then square root of, b, squared, minus, 4, a, c, end square root is a
**real number**, and the quadratic equation has**two real solutions**, start fraction, minus, b, minus, square root of, b, squared, minus, 4, a, c, end square root, divided by, 2, a, end fraction and start fraction, minus, b, plus, square root of, b, squared, minus, 4, a, c, end square root, divided by, 2, a, end fraction. - If b, squared, minus, 4, a, c, equals, 0, then square root of, b, squared, minus, 4, a, c, end square root is also 0, and the quadratic formula simplifies to start fraction, minus, b, divided by, 2, a, end fraction, which means the quadratic equation has
**one real solution**. - If b, squared, minus, 4, a, c, is less than, 0, then square root of, b, squared, minus, 4, a, c, end square root is an
**imaginary number**, which means the quadratic equation has**no real solutions**.

### Try it!

## Your turn!

## Things to remember

For a, x, squared, plus, b, x, plus, c, equals, 0:

- If b, squared, minus, 4, a, c, is greater than, 0, then the equation has 2 unique real solutions.
- If b, squared, minus, 4, a, c, equals, 0, then the equation has 1 unique real solution.
- If b, squared, minus, 4, a, c, is less than, 0, then the equation has no real solution.

## Want to join the conversation?

- What's a quick way to know if you should use quadratic formula or factorization method to solve a quadratic equation(3 votes)
- It largely depends on how good you are at factoring. Factoring is generally faster, but if it's a challenge to try and find the terms then the quadratic formula is more consistent. If when looking for factors you find yourself stuck for maybe more than 30 seconds, you might want to move onto the quadratic method, which you know will work every time. If there is no coefficient on the x^2 term (like in x^2 + 6x + 9), factoring is usually easier. Quadratics with "weird" coefficients (like radicals or decimals) might be better on your calculator in the quadratic formula than factoring by hand. It's really a case by case basis on which to choose, and as you practice more you'll develop that muscle.(3 votes)

- I think the answer to this problem is incorrect

x2+x−56=0x, squared, plus, x, minus, 56, equals, 0

What are the solutions to the equation above?(0 votes)- It's answer is correct.."x2-7x+8x-56" this the equation by which you get the factors. Thank you!(8 votes)

- -3x^2 + 12x + 4 = 0

This is the second last question, for this equation, for the answer given in in the explanation, why doesn't the four get square rooted as well when the 8 is divided in to 2*4(2 votes)- Try using the quadratic formula.(2 votes)

- What are the dots for in the square root?(1 vote)
- Multiplication(3 votes)

- how dose x=5 = to 0

and how dose 2/5 =0(1 vote)- If you're talking about the zero method, then I'm not 100% sure, but I think you just need to write it like x=-5 and x=-5/5. You need to add it to both sides of the equation (x+5=0 5-5=0 0-5=-5 x=-5). The whole part where you need to figure out which part makes it zero is just if you want to thoroughly check it.(1 vote)

- at13:26sal uses the numerator(2) to divide the denominator(-6), so how did -12 also reduce to -6 because it is really confusing me(1 vote)
- This is the distributive property of multiplication/division. Just like how (6x + 12y)/6 means that we need to divide both terms by 2 and end up with x + 2y, the same goes for the +/- in the quadratic formula.

Sal doesn't divide by 6, he instead divides by 2 in order to cancel out the coefficient on the square root. This means that the denominator becomes -3, and the -12 becomes -6.(1 vote)

- in the zero product property solutions, what do you do if both are simultaneously equal to zero

for example in (2x-1)(x+4)=0 --- what happens when both (2x+1) and (x+4) are zero at the same time? (since 0 x 0 will still be equal to 0)(1 vote) - “Which of the following values of x satisfy the equation -3x^2+12x+4=0?”

how the answer for this question has a positive denominator?

Cause in the whole equation he is negative, i didnt understand how it got positive(1 vote)- The negative sign in the denominator cancels out with the +/- sign. -*+=- and -*-=+, so you still have a negative and positive there, which corresponds to the plus-or-minus in the answer. Does that help?(1 vote)