It’s no secret that algebra is the most-tested math topic on the SAT. Fully 70% of the math questions are, directly or indirectly, algebra questions. Of course, the questions are not all “pure” algebra questions. Many of them are word problems that use algebra as a tool to determine the correct answer, and some may be questions from such diverse disciplines as geometry or statistics. Yet, algebra is the common thread.

In this article, we will focus on quadratic equations on the SAT, one of the most common algebra questions on the exam. We’ll cover solving quadratic equations, and although graphing quadratic functions is on the SAT, we will stick to analyzing quadratic functions in this article.

So What Is a Quadratic Equation?

The two most common algebra topics that you will encounter on the SAT are linear equations and quadratic equations. Linear equations contain a variable whose highest power is 1. An example of a linear equation is 3x + 5 = 11.

Quadratic equations contain a variable whose highest power is 2. Some examples of quadratic equations are:

y = 5x^2          y = x^2 + 5x + 6          0 = (x – 4)(x + 7)

Our goal in solving an algebraic quadratic equation is either to solve for the value(s) of x or to graph the quadratic function. Before we focus on solving these equations, let’s get comfortable manipulating them.

Keep in mind that you won’t have to spend too much time with questions where you are applying quadratics to real-world problems. Rather, the quadratic questions you see on the SAT will be pretty straightforward. Now, let’s get into FOILing quadratic equations.

KEY FACT:

A quadratic equation contains a variable whose highest power is 2.

The FOIL Method

You likely remember the FOIL method if you’ve taken an algebra class. But just in case you’re a bit rusty, let’s do a quick review of the technique. First, the name FOIL is a mnemonic for “First,” “Outside,” “Inside,” and “Last.” Let’s look at how we apply this mnemonic with an example problem.

How do we expand the quadratic equation y = (x + 3)(x – 5) ?

First, we multiply the “Firsts” of each factor in the expression. In our equation, we see that the “firsts” are x and x.

y = (x + 3)(x – 5)

When multiplying the “firsts,” we get x times x, or x^2.

Second, we multiply the “Outsides.” This means we multiply the two terms that are on the extremes.

y = (x + 3)(x – 5)

When we multiply the “Outsides,” we get x times -5, or -5x.

Third, we multiply the “Insides.” This means we multiply the two inner terms:

y = (x + 3)(x – 5)

When we multiply the “Insides,” we get 3 times x, or 3x.

Finally, we multiply the “Lasts,” which means we multiply the two “last” terms in each set of parentheses.

y = (x + 3)(x – 5)

When we multiply the “Lasts,” we get 3 times -5, or -15.

Combining the four steps of the FOIL process, we get:

y = (x + 3)(x – 5)  →  y = x^2 – 5x + 3x – 15  →  y = x^2 – 2x – 15

KEY FACT:

The FOIL method of expanding a quadratic equation in the form 0 = (x – a)(x – b) involves multiplying the “firsts,” the “outsides,” the “insides,” and the “lasts.”

Let’s now practice with an example.

Example 1: FOIL

Which of the following is equivalent to the quadratic equation y = (2x – 5)(x + 3) ?

• (A) y = 3x^2 – 11x – 15
• (B) y = 2x^2 – 11x – 15
• (C) y = 2x^2 + x – 15
• (D) y = 2x^2 – 8x + 15

Solution:

Let’s use the FOIL method to expand y = (2x – 5)(x + 3).

Firsts:  (2x)(x) = 2x^2

Outsides: (2x)(3) = 6x

Insides: (-5)((x) = -5x

Lasts: (-5)(3) = -15

We combine these to obtain:

2x^2 + 6x – 5x – 15

2x^2 + x – 15

Let’s now consider the opposite operation of FOILing: factoring.

The general form of a quadratic is the trinomial form ax^2 + bx + c = 0. In this form, the constant term c is also the y-intercept of the quadratic when it is graphed. We call the equation a trinomial because it has three terms:  ax^2, bx, and c.

As such, it may be useful to reorganize the equation into two binomial factors. In other words, if we have the trinomial x^2 -2x – 15 = 0, it might be advantageous to re-express it as the product of two binomial factors: (x + 3)(x – 5) = 0.

(If you are wondering why quadratic equations break down to two binomial factors, remember that when graphed, a quadratic equation is in the form of a parabola, which usually has two x-intercepts.)

So how do we go “backward” from a trinomial to a product of two binomial factors? The technique is called “factoring,” and we will look at the steps involved.

The Four-Step Procedure for Factoring a Quadratic Trinomial

Let’s consider factoring the trinomial x^2 – 3x = 10.

Step 1: First, we put the quadratic equation in standard form  ax^2 + bx + c = 0.

x^2 – 3x = 10  →  x^2 – 3x – 10 = 0

Step 2: Identify the coefficients a and b and the constant term c.

a = 1 because the coefficient of x^2 is 1.

b = -3 because the coefficient of -3x is -3.

c = -10 because the constant term is -10.

Step 3: If a = 1 (which is the case in most SAT quadratic questions), find two numbers – let’s call them p and q – such that they have a sum of b and a product of c.

Here, we want to find two numbers (p and q) such that their sum is -3 and their product is -10. Often, you have to do this step using trial and error. Let’s first look at the numbers that have a product of -10:

1 and -10        -1 and 10       2 and -5        -2 and 5

Looking at our options, we see that only one of these pairs also has a sum of -3. Thus, the two numbers are p = 2 and q = -5.

Step 4: We now re-express the trinomial as a product of two binomial factors in the form 0 = (x + p)(x + q)

Our trinomial gave us values for p and q of 2 and -5, respectively. Thus, we can re-express the trinomial x^2 – 3x – 10 =  0 in factored form as:

0 = (x + 2)(x – 5)

KEY FACT:

Let’s try an example.

What is the correct factoring of the quadratic equation x^2 + 5x + 6 = 0?

• (x + 3)(x + 2) = 0
• (x – 3)(x – 2) = 0
• (x + 3)(x – 2) = 0
• (x – 3)(x + 2) = 0

Solution:

Let’s factor the quadratic equation by using the four steps presented earlier.

Step 1: We see that x^2 + 5x + 6 = 0 is in the correct form, ax^2 + bx + c = 0.

Step 2: For x^2 + 5x + 6 = 0, we see that a = 1, b = 5, and c = 6.

Step 3: We see that a = 1, so we proceed. First, we find the two numbers – p and q – that have a product of c = 6 and a sum of b = 5. The pairs of numbers with a product of 6 are:

1 and 6        -1 and -6         2 and 3        -2 and -3

Of the pairs of numbers with a product of 6, only the numbers 2 and 3 also produce a sum of 5. Therefore, the values of p and q are 2 and 3, respectively.

Step 4: We now re-express the trinomial x^2 + 5x + 6 = 0 as a product of two binomial factors in the form (x + p)(x + q) = 0:

(x + 2)(x + 3) = 0

The Zero Product Property

In this lesson and the ones that follow, we will focus on finding solutions to quadratic equations. Along the way, we will use the skills we just learned for combining and simplifying quadratic expressions. We will also frequently make use of a mathematical fact called the zero product property.

The zero product property states that if the product of two quantities is equal to 0, then at least one of the quantities must be equal to 0.

For example, if a x  b = 0, then one of the following scenarios must be true:

Scenario 1: a = 0

In this case, b can equal any number. We can confirm this with actual numbers:

If a = 0 and b = 4, we see that 0 x 4 = 0.

If a = 0 and b = -20, we see that 0 x (-20) = 0.

Scenario 2: b = 0

In this case, a can equal any number. For example:

If a = 6 and b = 0, we see that 6 x  0 = 0.

If a = 3.5 and b = 0, we see that 3.5 x 0 = 0.

Note: It is also possible that both a and b are zero, since 0 x 0 = 0.

We may also find applications of the zero product property when dealing with just one variable. For example, if (n + 4)(n – 1) = 0, then one of the following scenarios must be true:

Scenario 1: n + 4 = 0

Solving this equation for n, we find that n = -4. Substituting n = -4 back into the original equation, we see that the product will be zero:

(n + 4)(n – 1)   →  (-4 + 4)(-4 – 1)   →  (0)(-5) = 0

Scenario 2: n – 1 = 0

Solving this equation for n, we find that n = 1. Substituting n = 1 back into the original equation, we see that the product will be zero in this scenario as well:

(n + 4)(n – 1)   →  (1 + 4)(1 – 1)   →  (5)(0) = 0

Thus, we can conclude that either n = -4 or n = 1. Those are the only two solutions of this equation.

KEY FACT:

The zero product property states that if two quantities multiply to zero, then at least one of those quantities must be equal to zero.

Let’s consider the following examples.

Example 3: Zero Product Property

Which of the following represents the set of possible solutions for x in the quadratic equation

0 = x^2 – 7x + 12

• {4}
• {3, 4}
• {-3, -4}
• {3}

Solution:

First, let’s factor the trinomial into its two binomial factors:

Step 1: We see that x^2 – 7x + 12 = 0 is in the correct form, ax^2 + bx + c = 0.

Step 2: For x^2 – 7x + 12 = 0, we see that a = 1, b = -7, and c = 12.

Step 3: We see that a = 1, so we proceed. First, we find the two numbers – p and q – that have a product of c = 12 and a sum of b = -7. The pairs of numbers with a product of 12 are:

1 and 12        -1 and -12         2 and 6        -2 and -6     3 and 4     -3 and -4

Of the pairs of numbers with a product of 12, only the numbers -3 and -4 also produce a sum of -7. Therefore, the values of p and q are -3 and -4, respectively.

Step 4: We now re-express the trinomial x^2 -7x + 12 = 0 as a product of two binomial factors in the form (x + p)(x + q) = 0:

(x – 3)(x – 4) = 0

We now use the zero product property to find the values of x that satisfy the equation (x – 3)(x – 4) = 0.

x – 3 = 0    →   x = 3

x – 4 = 0     →  x = 4

We see that either x = 3 or x = 4 satisfies the equation.

On the SAT, three identities are so commonly used that memorizing them is recommended.

(x + y)^2 = (x + y)(x + y) = x^2 + 2xy + y^2

(x – y)^2 =  (x – y)(x – y) = x^2 – 2xy + y^2

(x – y)(x + y) = x^2 – y^2

TTP PRO TIP:

Memorize the three common quadratic identities:

(x + y)^2 = x^2 + 2xy + y^2

(x – y)^2 =  x^2 -2xy + y^2

(x – y)(x + y) = x^2 – y^2

The third quadratic identity (x – y)(x + y) = x^2 – y^2 deserves special attention. It is called the difference of squares, and you will see it quite often in the Math section of the SAT. Let’s look at an example that uses the difference of squares so that you can see how it might appear in an SAT question.

If x ≠ 6, then what value of x satisfies the equation

(x^2 – 36) / (x – 6) = 5 ?

First, we recognize that x^2 – 36 matches the pattern of the difference of squares: x^2 – y^2. We can factor x^2 – 36 as (x – 6)(x + 6).Thus, we have:

(x^2 – 36) / (x – 6) = 5

(x – 6)(x+ 6) / (x – 6) = 5

We can cancel (x – 6) from both the numerator and denominator, obtaining:

x + 6 = 5

x = -1

By using this special quadratic identity, we saved time and effort in determining the solution to the equation in the question stem.

Let’s look at another example that uses another one of the quadratic identities.

Which of the following is equivalent to [3x^2 – 10x + 21 – (-4 + 2x^2)] / (x – 5) for x ≠ 5?

• x – 25
• 2x – 5
• x – 5
• x + 5

Solution:

First, let’s simplify the given expression.

[3x^2 – 10x + 21 – (-4 + 2x^2)] / (x – 5)

[3x^2 – 10x + 21 + 4 – 2x^2] / (x – 5)

(x^2 – 10x + 25) / (x – 5)

The numerator is in the form x^2 – 2xy + y^2, where x = x and y = 5. This means we can factor the expression using the quadratic identity x^2 – 2xy + y^2 = (x – y)^2. The numerator factors to (x – 5)^2, giving us:

(x – 5)^2 / (x – 5)

(x – 5)

For our final topic, let’s study one of the most useful formulas in all of algebra – the quadratic formula.

Throughout this article, we have limited our discussion of quadratic equations in the form ax^2 + bx + c = 0 to equations where a = 1. This has made the factoring and the solution of the equations relatively easy. Now, we will consider quadratic equations in which the value of the coefficient a might not equal 1. While there are many techniques for factoring and solving for the value(s) of x in this situation, the most common way is to use the quadratic formula. Let’s see what it is and how to use it!

Consider the following quadratic equation:  x^2 – 2x – 8 = 0. We can factor this, using techniques we covered earlier in this article and noting that a = 1, b = -2, and c = -8, as (x – 4)(x + 2) = 0. Using the zero product property, we see that x = 4 or x = -2.

Any quadratic equation in the form ax^2 + bx + c = 0 can be solved by using the quadratic formula:

x = [-b +/- √(b^2 – 4ac)] / 2a

To practice using the quadratic formula, let’s use the example x^2 – 2x – 8 = 0, which we previously showed as having solutions of x = 4 or x = -2.

Let’s substitute the values a = 1, b = -2, and c = -8 into the quadratic formula:

x = [-b +/- √(b^2 – 4ac)] / 2a

x = [-(-2) +/- √((-2)^2 – 4(1)(-8))] / 2(1)

x = [2 +/- √(4 +32)] / 2

x = [2 +/- √(36)] / 2

x = (2 +/- 6) / 2

x = (2 + 6) / 2    or   x =  (2 – 6) / 2

x = 4    or    x = -2

Notice that the same solutions (x = 4 or x = -2) were found using either method. The good news is that the quadratic formula gives you the solutions of any quadratic equation, no matter how convoluted it looks. Let’s look at an example.

Consider the quadratic equation 0 = 3x^2 -7x + 2. We know that a = 3, b = -7, and c = 2. Let’s substitute those values into the quadratic formula and solve for x.

x = [-b +/- √(b^2 – 4ac)] / 2a

x = [-(-7) +/- √((-7)^2 – 4(3)(2))] / 2(3)

x = [7 +/- √(49 – 24)] / 6

x = [7 +/- √(25)] / 6

x = (7 +/- 5) / 6

x = 12/6    or   x =   2/6

x = 2     or    x = 1/3

Let’s try an example.

What is the sum of all solutions to the quadratic equation 3x^2 – 12x + 8 = 0 ?

• (12 + 4√3) / 3
• (12 + 4√15) / 3
• 4
• 8

Solution:

The quadratic equation 3x^2 – 12x + 8 = 0 has a = 3, b = -12, and c = 8. Plugging these values into the quadratic formula, we get:

x = [-b +/- √(b^2 – 4ac)] / 2a

x = [-(-12) +/- √((-12)^2 – 4(3)(8))] / 2(3)

x = [12 +/- √(144 – 96)] / 6

x = [12 +/- √(48)] / 6

x = (12 +/- 4√3) / 6

x = 6 +/- 2√3) / 3

x = (6 + 2√3) / 3   or   x = (6 – 2√3) / 3

Now we sum the two roots:

(6 + 2√3) / 3  +  (6 – 2√3) / 3

12/3 = 4

Summary

A quadratic equation contains a variable whose highest power is 2. An example is y = 5x^2 + 12x + 16. Quadratic equations appear quite often on the SAT, so familiarity with them is paramount.

• The FOIL method: (x – 3)(x + 7) = x^2 + 7x – 3x – 21 = x^2 + 4x – 21.
• Factoring: x^2 + 7x + 10 = (x + 5)(x + 2)
• The Zero Products Property:  If (x – 4)(x + 5) = 0, then x = 4 or x = -5