GRE Exam: A Simple Guide To Algebra - Quadratic Equations

Quadratic Equations
So far, you have looked only at linear equations. In linear equations, there will always be one solution for a given variable. In contrast, in quadratic equations, a variable will usually have more than one solution.

How do you know you have a quadratic equation?
You have a quadratic equation whenever at least one of the variables in the equation is raised to an even exponent.

Let’s say you are asked to evaluate (−4)2. Using PEMDAS, you know that the result is (−4) × (−4) = 16.

Now let’s say that you are asked to evaluate 42. You get 4 × 4 = 16.

Note that both 4 and −4 gave the same result.

Why? Because raising a variable to an even exponent will always produce a positive result.

Now let’s flip it:
If x2 = 16, what are the possible values of x?

You may be tempted to calculate the square root of 16 and say that x = 4.

But watch out for the even exponent!

Since the exponent on x is even, there will be a positive and a negative solution for x.

Thus the two solutions are x = 4 or −4.

Other forms of quadratic equations are:


Common Templates of Quadratic Equations
On the GRE, quadratic equations will take a few common forms. The most common form is

When presented with a quadratic equation in the preceding form, you will usually be asked to find the solutions of that equation. To do so, you will need to factor.

Let’s look at an example:

Step 1: Manipulate the equation to match the preceding template. In this example, you would need to set the equation equal to zero.

Step 2: Rewrite the equation in factored form: x2 + 7x + 12 = (x + __ )(x + __).

Step 3: Determine the values for the slots. To get the factors of the equation, you need to find two integer values that add to yield your b and that multiply to yield your c. In the preceding equation, 7 is your b and 12 is your c. What two values multiply to 12 and add to 7? 3 and 4.
In the slots, you will put the two integer values that you arrived at in Step 2. Thus in its factored form, the equation is (x + 3)(x + 4) = 0.

Step 4: Solve for x. To solve for x, you must recognize an essential fact: any time a product of two or more factors is zero, at least one of those factors must have a value of zero.

In the preceding example, if (x + 3)(x + 4) = 0, then either:

So the roots of this equation are −3 and −4.

Note that if you plug either of these values into the original equation, you will arrive at a true statement.

Setting the Quadratic Equation Equal to Zero
Oftentimes, you will be presented with a quadratic equation that does not appear to match the preceding template.

For example, if 4x2 = x, then x = ?

Seeing x on both sides of the equation, many students are tempted to divide both sides of the equation by x to arrive at:

Though is certainly a solution to the equation, the hypothetical student committed an error here when dividing by x.

Why?

Because the student essentially eliminated one of the solutions for x! Instead of arriving at two solutions, the student arrived at only one.

So stick to the following rule: Set quadratic equations equal to zero. Let’s redo the preceding example:

Step 1: Subtract x from both sides: 4x2 − x = 0.
Step 2: Factor x from both terms: x(4x − 1) = 0.
Step 3: Solve for x.

Since you have a product set equal to zero, either:


Expanding a Quadratic: FOIL
So far, you have looked at situations where you have taken quadratic equations in their expanded form and put them into factored form. Sometimes, you will be expected to go in the opposite direction: from factored form to expanded form.

To do so, you will want to use an acronym that you may remember from high school: FOIL.

FOIL stands for:
First
Outer
Inner
Last

To expand the expression (x + 3)(x − 5), do the following:
First: Multiply the first term in each parentheses together: (x)(x) = x2
Outer: Multiply the first term in the first parentheses by the last term in the second parentheses: x(−5) = −5x
Inner: Multiply the inner terms of the product together: 3(x) = 3x
Last: Multiply the last term in each set of parentheses together: 3(5) = –15

Now you have an expression with four terms: x2 − 5x + 3x + 15. Group like terms, and you will arrive at the quadratic: x2 − 2x + 15.

When you have the opportunity to factor or use FOIL on the GRE, it’s usually a good idea to do so!

Common Quadratics
Three quadratic expressions appear so frequently on the GRE that it is worth memorizing their structure instead of factoring or using FOIL each time you encounter them:
1. (x + y)(x − y) = x2 − y2
2. (x + y)(x + y) = (x + y)2 = x2 + 2xy + y2
3. (x − y)(x − y) = (x − y)2 = x2 − 2xy + y2

Memorizing the preceding formulas is useful for a couple of reasons:
- If you know the forms of these expressions after applying FOIL and after factoring, you will be able to save time when you encounter their general form on the GRE.

- Often, the GRE will put these expressions in an unorthodox form. In these cases, you will need to recognize that an unusual-seeming expression is actually one of the common quadratics.

For example,
Since you are multiplying two binomials, you might be tempted to distribute, but notice that is in the same form as (x + y)(x − y), where is x and is y.

From the first special product, you know that (x + y)(x − y) = (x2 − y2). Therefore,