A Guide to the GRE/Quadratic Equations
Contents
Quadratic Equations[edit]
A quadratic equation is an equation such as x^{2}  3x = 2, where x is multiplied by both itself and a constant.
Rule[edit]
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}  3x = 3




 Start with the initial equation.



x^{2} 3x + 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.



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]
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]
“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]
1. If x^{2} + 5 = 6x, then what is the value of x?
2. x^{2}  2x = 8. Solve for x.
3. If x^{2} + 4x + 2 = 1, then what does x equal?
Answers to Practice Questions[edit]
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.