# A Guide to the GRE/Quadratic Equations

# Quadratic Equations[edit | edit source]

A **quadratic equation** is an equation such as *x*^{2} - 3*x* = 2, where *x* is multiplied by both itself and a constant.

## Rule[edit | edit source]

**To solve an equation such as “ x^{2} + 2x = -1”, adjust the equation so that a zero is on one side, then break the equation into factors such as (x + 1)(x + 1) = 0. The correct answer is the opposite of the numbers in the parentheses.**

*x*^{2} - 3*x* = -3

- Start with the initial equation.

*x*^{2} -3*x* + 3 = 0

- Add 3 to both sides to set the equation equal to zero.

(*x* + ?)(*x* + ?) = 0

- Break the equation into two “roots” or “factors.”

(*x* - 3)(*x* - 1) = 0

- Find the two numbers which add up to the second number (the -3) and multiply to the third number (the 3). These will be the numbers in the parentheses, that multiply to
*x*in the original equation.

- Find the two numbers which add up to the second number (the -3) and multiply to the third number (the 3). These will be the numbers in the parentheses, that multiply to

*x* = 3, 1

Most quadratic equations have two answers because they have different numbers in each factor. Some, however, like (x + 1)(x + 1), only have one answer.

### Difference of Two Squares[edit | edit source]

*x*^{2} - *y*^{2} factors out to (x - y)(x + y). This is known as the rule of “difference of two squares.”

Thus, *x*^{2} - 9 would thus factor out to (*x* - 3)(*x* + 3). This is a special rule that is often tested on the GRE.

### Factoring[edit | edit source]

“Factoring” is one of the most difficult parts of solving a quadratic equation. x^{2} + 6x + 9 = 0 will factor into (x + 3)(x + 3), because if multiplied out, (x + 3)(x + 3) equals the prior equation. The numbers in the parentheses will add up to second number in the equation x^{2} + 6x + 9 = 0 and multiply up to the third number. Observe more examples:

Original equation Set it equal to zero Factor Solution

x^{2} + 4 = -4x x^{2} + 4x + 4 = 0 (x + 2)(x + 2) x = -2

x^{2} -4x = -3 x^{2} - 4x + 3 = 0 (x - 3)(x - 1) x = 3, 1

x^{2} - 1 = x x^{2} - x - 2 = 0 (x - 2)(x + 1) x = 2, -1

(find two numbers that add to the middle number and multiply to the third number)

## Practice[edit | edit source]

1. If *x*^{2} + 5 = -6*x*, then what is the value of x?

2. *x*^{2} - 2*x* = 8. Solve for *x*.

3. If *x*^{2} + 4*x* + 2 = -1, then what does *x* equal?

## Answers to Practice Questions[edit | edit source]

1. -5, -1

x2 + 5 = -6x Take the initial equation.

x2 + 6x + 5 = 0 Add 6x to both sides to set the equation equal to zero. Make sure that the equation is in the form x2 + bx + c = 0.

(x + ?)(x + ?) Break the equation into factors. What two numbers add up to the middle number (6) and multiply to the last number (5)?

(x + 5)(x + 1) = 0 Plug the correct numbers into the equation.

x = -5, -1 x is the opposite of these numbers.

2. 4, -2

x2 - 2x = 8 Take the initial equation.

x -2x - 8 = 0 Subtract 8 from both sides to set the equation equal to zero. Make sure that the equation is in the form x2 + bx + c = 0.

(x + ?)(x + ?) Break the equation into factors. What two numbers add up to the middle number (-2) and multiply to the last number (-8)?

(x - 4)(x + 2) = 0 Plug the correct numbers into the equation.

x = 4, -2 x is the opposite of these numbers.

3. -3, -1

x2 + 4x + 2 = -1 Take the initial equation.

x2 + 4x + 3 = 0 Add 1 to both sides to set the equation equal to zero. Make sure that the equation is in the form x2 + bx + c = 0.

(x + ?)(x + ?) Break the equation into factors. What two numbers add up to the middle number (4) and multiply to the last number (3)?

(x + 3)(x + 1) = 0 Plug the correct numbers into the equation.

x = -3, -1 x is the opposite of these numbers.

Some quadratic equations cannot be factored and must be solved using a special formula, the “quadratic formula.” This formula is not tested on the GRE.