Math Study Guide for the SAT Exam

Advanced Math

These questions require you to use and manipulate expressions and equations as well as to interpret and build functions. Additionally, the equations with which you work on this type of question tend to be more complex.

Look through these topics and be sure you know all their ins and outs. Since these topics can be difficult, doing practice questions for anything in this section is highly recommended.

Equivalent Expressions

Equivalent expressions are expressions that look different but actually represent the same value. Using equivalent expressions, you can simplify problems, check your work, and solve equations. The process often involves the techniques you’ve learned for manipulating algebraic terms, such as factoring, distributing, combining like terms, and rearranging variables. To find if two expressions are equivalent, you can perform the same mathematical operation on both or simplify each side separately to see if they reduce to the same result. For example, \(2(x+3)\) and \(2x +6\) are equivalent because distributing the \(2\) in the first expression produces the second expression. Similarly, \(3a - 2a + 5\) simplifies to \(a + 5\), which means any longer expression that simplifies to \(a + 5\) is also equivalent. Just note that equivalent expressions always yield the same value for all allowable inputs for the variable.

Nonlinear Equations in One Variable

When an equation involves powers, roots, or other operations beyond the first degree, it is called a nonlinear equation. Unlike linear equations, which present as a straight line, nonlinear equations curve upward, downward, or create other shapes when graphed.

A nonlinear equation in one variable can take many forms, including those with radicals, squares (quadratics), or variables in the denominators. Each has its own method for solving, but the strategy is the same: You use algebra to isolate the variable. For roots, it’s always important to check your final answer, since these often produce extraneous solutions. In the sections below, you will learn how to handle radical equations, quadratics, and equations with variables in the denominator.

Radical Equations

If we’re faced with an equation involving a radical, we need to get rid of it. We do this by raising each side of the equation to the power corresponding to the root in the radical. For example, if we have a square root in an equation, we need to square that square root so we’re left with the radicand (the number inside the radical symbol).

We’ll do an example by solving for \(x\) in this equation:

\[3 = 1 + \sqrt{(x-5)}\]

If we square it right now, when we multiply out everything on the right side, we’ll have:

\[1 + 2\sqrt{(x-5)}+(x-5)\]

However, we will still be stuck with the radical. So, we need to isolate the radical first. To do so, we’ll subtract \(1\) from each side to get:

\[2 = \sqrt{(x-5)}\]

Now we can get rid of the root by squaring both sides:

\[4 = x - 5\] \[x = 9\]

We can plug that back in and check that the equation holds. It’s a good idea to check because we can’t just “undo” squaring. There are both positive and negative roots.

Quadratic Equations

Quadratic equations are a specific type of polynomial equation that only go up to the second power of their variable, so the most general form is \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are any constant numbers. We give them a special name because they appear more frequently and are easy to analyze.

Methods of Solving

Suppose we’re given a quadratic equation in the form \(ax^2 + bx + c = 0\) and are asked to solve it. There are several different ways to think about and solve this problem. Let’s look at each of them.

Factoring Quadratic Equations

If you can identify the factors, factoring is the fastest method. That is, if we can rewrite \(ax^2 + bx + c\) as a product of two terms, \((x + s)(x + t)\), then when we set this equal to zero, we see that the only way for the equation \((x + s)(x + t) = 0\) to be true is if one of the terms equals zero. In other words, either \(x + s = 0\) or \(x + t = 0\). This allows us to quickly see the solutions must be \(x = -s\) and \(x = -t\).

Therefore, to factor the equation, first, we divide each term by \(a\) to get the \(x^2\) on its own. Once that is done, we can factor the quadratic as the following:

\[x^2 + \frac{b}{a}x + \frac{c}{a} = (x + s)(x + t)\]

It can be expanded to:

\[x^2 + \frac{b}{a}x + \frac{c}{a} = x^2 + (s + t)x + st\]

In other words, if we can find two terms \(s\) and \(t\) so that \(\frac{b}{a} = s + t\) and \(\frac{c}{a} = st\), then we can factor the original equation, \(ax^2 + bx + c = 0\), as:

\[a(x-s)(x-t) = 0\]

We can’t always factor quadratic equations with real numbers, though, and even when we can they might be difficult to mentally calculate, so it is wise to have a more general approach as a backup. Fortunately, we have another trick to use on quadratic equations.

Factoring Higher Order Polynomials

When we move beyond quadratic equations to polynomials of three degrees (cubic) or higher, the overall goal remains the same. We rewrite the expression as a product of simpler factors so we can use the zero-product property to find solutions. However, higher-degree polynomials often require additional strategies because they may not factor as neatly as quadratics.

A common first step is to check whether the polynomial has a greatest common factor (GCF). If so, factor it out immediately. After that, there are several techniques that frequently help such as:

• Factoring by grouping—This method is useful when the polynomial has four or more terms arranged in a way that allows you to factor pairs of terms.

• Recognizing special patterns—Certain patterns, such as perfect cube formulas or the difference of cubes, can provide shortcuts.

• Using known or testable roots—This method can be used to find a factor of the form \(x - r\). Once one factor is identified, the polynomial can be divided by that factor to reduce the degree and then factored further.

Let’s take a look at an example. Consider this equation:

\[x^3 - 4x^2 - x + 4 = 0\]

We can start to factor by grouping:

\[(x^3 - 4x^2) + (-x + 4)\]

Now, we’ll factor each pair:

\[x^2(x - 4) - 1(x - 4)\]

Now the expression has a common factor of \(x - 4\), so we can write it as:

\[(x - 4)(x^2 - 1)\]

The remaining quadratic can be factored further using the difference of squares:

\[(x - 4)(x - 1)(x + 1)\]

Thus, the solutions to the original equation are:

\[x = 4,\quad x = 1,\quad x = -1\]

Factoring higher-degree polynomials takes practice, but the key idea is always the same. You look for structure, apply known patterns, and reduce the polynomial step by step until it can be solved using familiar methods.

Equations with Variables in the Denominator

When faced with an equation with variables in the denominators of some terms, the most helpful approach is to multiply by the denominator to cancel them out and leave everything in terms of polynomials, which we can then solve with the techniques we discussed earlier. For example, consider this equation:

\[\frac{5}{b+1} = \frac{3 - 1}{b+2}\]

When we multiply by \((b+1)(b+2)\), the equation becomes:

\[5(b + 2) = (3 - 1)(b + 1)\] \[5b + 10 = 3b + 3 - b - 1\] \[5b + 10 = 2b + 2\]

Now, solving for \(b\), we get:

\[3b = -8\] \[b = \frac{-8}{3}\]

Systems of Nonlinear Equations in Two Variables

We solve a system of two linear equations with the substitution or elimination techniques you learned earlier. Try to isolate one of the variables so you can solve for its value. If you can isolate a variable, substitution works the same way as before. For example, let’s solve this system:

\[\begin{Bmatrix} x + y = 4\\ (y-1)^2 + xy = 1 \end{Bmatrix}\]

Using the first equation, we can isolate \(x = 4 - y\) and substitute this into the other:

\[1=(y-1)^2 + xy = (y-1)^2 + (4-y)y\] \[= y^2 - 2y + 1 + 4y - y^2 = 2y + 1\]

Now, we have \(2y + 1 = 1\), which means that \(y = 0\).

As such, \(x = 4 - y = 4\), so the solution is \((x, y) = (4, 0)\).

At times, isolating a variable for the substitution might not be possible or easy. In such a case, you can substitute the form of the equations. Consider this system:

\[\begin{Bmatrix} 5y + 2 = (x - 8)^2 - x\\ x = 2(y + 4) \end{Bmatrix}\]

The \(x\) is already isolated in the second equation, but to use it we would need to expand \((x-8)^2\). It would be faster to notice that \(x=2(y+4)\) can be rewritten as \(x - 8 = 2y\) (and \(x=2y+8\)). Therefore, we can insert both into the other equation as such:

\[5y + 2 = (x - 8)^2 - x\] \[5y + 2 = (2y)^2 - (2y + 8) = 4y^2 - 2y - 8\]

So:

\[0 = 4y^2 - 7y - 10\]

We still need to solve a quadratic equation for \(y\) and then use the linear equation to find \(x\), but this is a shortcut that avoids having to expand the square.

Systems of Equations in the Real World

On the SAT, you will have word problems that present two distinct pieces of information that can be represented with variables. In such cases, you may find the best way to solve the question is to create a system of equations. Let’s try an example problem.

Pens cost \(50\) cents and pencils cost \(25\) cents. Amanda purchased a total of nine pens and pencils for a cost of \(\$4.25\). What is the number of pens she bought and what is the number of pencils she bought?

Solution

We have been given two pieces of information: the total number of pens and pencils and the prices of each. Both the associated costs and the total amount are fixed values, but the number of pens and pencils is unknown. Those are our variables. Let’s use \(x\) to represent the number of pens and \(y\) to represent the number of pencils. We can generate a system of equations as follows:

\[\begin{Bmatrix} 50x + 25y = 425\\ x + y = 9 \end{Bmatrix}\]

Solving the second equation for \(y\) and substituting the expression into the first equation, we get:

\[y = 9 - x\] \[50x + 25(9 - x) = 425\] \[50x + 225 - 25x = 425\] \[25x = 200\] \[x = 8\]

Finally, substituting this value into the second equation gives us:

\[8 + y = 9\] \[y = 1\]

Therefore, Amanda bought eight pens and one pencil.