# SAT Tip: Algebra Concepts You Need to Know

*This tip on improving your SAT score was provided by Vivian Kerr at Veritas Prep.*

Solving for variables and doing basic arithmetic is a requirement for the SAT, but medium to harder-level questions will also require you to review quadratics and systems of equations and to deal with functions. Here’s a little “cheat sheet” review of these more advanced algebra ideas.

**Quadratics**

Quadratic equations are special equations that have three terms and are in the form ax^{2} + bx + c = 0.

An example of a quadratic is x^{2} – 6x + 9 = 0. Every quadratic has two factors that multiplied together equal the quadratic. Here’s an example:

Factors: (x – 6)(x + 2)

Quadratic: x^{2} – 4x – 12

How can we find the factors of any quadratic equation? There are three rules to remember to go from any quadratic equation to its factors. The following examples are for the quadratic x^{2} – 6x + 9 = 0.

**Rule 1:** The first term of each factor will be the square root of the first term in the quadratic. The first term in each factor of x^{2} – 6x + 9 = 0 must be x, since the square root of x^{2} is x.

**Rule 2:** The second terms in the two factors must add together to equal the middle term’s coefficient. The coefficient (the number in front of the variable) of the second term is -6. It’s important to include the sign in front of the integer as part of the coefficient.

Now we need to think of two numbers that add together to give us -6. A few possible pairs would work (-1 and 6, -2 and 3, -6 and 1, 2, and -3, etc.), so before we pick, let’s examine the third term of the quadratic. Here, it’s + 9.

**Rule 3:** The third term of the quadratic equation will equal the product of the second terms in the two factors.

So not only do we need the two numbers to add together to equal -6, but we also need them to multiply together to equal -6. Therefore the factors must be: (x – 3) (x – 3). The “roots” or the “solutions” for this quadratic would be 3 only, since x – 3 = 0 is solved by adding 3 to both sides.

**Systems of Equations**

The SAT will often present you with two or more equations with multiple variables. Remember the “n equations with n variables rule.” If there are two variables in an equation (for example, z and y), there usually (but not always) needs to be two equations that each contain those variables in order to solve.

If we were given two equations: x – 3y + 7z = 16, and x + 3y = 29, can we solve for all three variables? The answer is “no” because we have 3 variables (x, y, and z) but only two equations. We’d need one more equation to solve.

The two common ways to solve are “substitution” and “combination.” Look for future blogs that delve into each method in more detail.

**Functions**

Functions may look intimidating, but they’re just a fancy way of writing an equation. It’s helpful to think of (x, f(x)) as another way of writing (x, y). Instead of an equation such as y = 5x, a function looks like f(x) = 5x. The math isn’t too different. For many function questions, you can often pick numbers for the variables to solve.

1. For which of the following functions *f* is *f*(x) = *f*(1 – x) for all x?

(A) *f*(x) = 1 – x

(B) *f*(x) = 1 – x^{2}

(C) *f*(x) = x^{2} – (1 – x)^{2}

(D) *f*(x) = x^{2}(1 – x)^{2}

(E) *f*(x) = x / 1-x

For instance, here *f*(x) = *f*(1-x), so if we pick x = 4 and plug it into the equation, we can write the function as *f*(4) = *f*(-3). For each answer choice, we could replace each function with 4 and -3 and see if *f*(4) = *f*(-3). The answer is (D):

*f*(-3) = (-3)^{2} (1 – (-3)^{2} = (9)(16)

*f*(4) = (4)^{2} (1 – (4))^{2} = (16)(9)

*f*(-3) = *f*(4)

Algebra is the backbone of the SAT Math section. Keep practicing these concepts, and you’ll be set on Test Day.

*Vivian Kerr has been teaching and tutoring in the Los Angeles area since 2005. She graduated from the University of Southern California, studied abroad in London, and has worked for several test-prep giants tutoring, writing content, and blogging about all things SAT, ACT, GRE, and GMAT.*