4. Quadratic Equations

Question 1

What is a Quadratic Equation?

Answer 1

Quadratic equations are equations with one unknown and two defining components:

(1) a variable term raised to the second power
(2) a variable term raised to the first power

Quadratic equations usually have two solutions (i.e. there are usually two possible values of the variable that make the equation true)

Examples:
x^2 + 3x + 8 = 12
a = 5a^2

Question 2

What is FOIL?

Answer 2

(a + b)(x + y)
In order to distribute the product of two factors, multiply every term in the first sum by every term in the second sum, then add all the products up

First, Outer, Inner, Last

(1) Multiply the First term in each of the parentheses; a * x = ax
(2) Multiply the Outer term in each; a * y = ay
(3) Multiply the Inner term in each; b * x = bx
(4) Multiply the Last terms in each; b * y = by

Then add up the product (a + b)(x + y) = ax + ay + bx + by

Example:
(x + 5)(x – 4)
= x^2 – 4x + 5x – 20
= x^2 + x – 20

Question 3

Solve for w: 3w^2 = 6w

Answer 3

w(3w – 6) = 0
3w = 6
w = 2

w = 2 OR w = 0

Question 4

Solve for x: x^3 + 2x^2 – 3x = 0

Answer 4

x(x^2 + 2x – 3) = 0
x(x + 3)(x – 1) = 0

x = 0 OR x = -3 OR x = 1

Question 5

What is implied if you have a quadratic expression equal to 0 and you can factor an x out of the expression?

Answer 5

Then x = 0 is one solution of the equation

NOTE: DO NOT divide both sides by x. This division improperly eliminates the solution x = 0. You are only allowed to divide by a variable (or ANY expression) if you are absolutely sure that the variable or expression does not equal zero

Question 6

What is a perfect square quadratic?

Answer 6

A quadratic equation that can be solved without setting one side equal to zero. The problem can be quickly solved by taking the square root of both sides of the equation

Example:
(z + 3)^2 = 25
z + 3 = Sqrt(25)
z = 2 OR z = -8

Question 7

What are examples of one-solution quadratic equations?

Answer 7

x^2 + 8x + 16 = 0
(x – 3)^2 = 0
x = 3

(x^2 + x – 12) / (x – 2) = 0
(x – 3)(x + 4) / (x – 2) = 0
x = 3 OR x = -4
*Recall that if zero appears in the denominator, the expression becomes undefined. Thus, x = 2 is not a solution

Question 8

How do you factor a quadratic expression?

Answer 8

(1) Move all of the terms to one side of the equations, put them in the form of ax^2 + bx + c (where a, b and c are integers), and set them equal to zero
(2) Factor the equation by rewriting the distributed expression as a product of two terms (i.e. put the expression in its factored form)
(3) Solve for the value(s) of x

Question 9

How do you handle a positive constant in the quadratic expression?

Answer 9

The two numbers in the factored form must both be positive or both be negative

Question 10

How do you handle a negative constant in the quadratic expression?

Answer 10

Test factor pairs by pretending the constant is positive and asking which pair differs by the right amount

If the constant is negative, then one number in the factored form is positive, and the other one is negative

Question 11

How do you handle a negative x^2 term?

Answer 11

Factor out a common factor of -1 first

Question 12

How do you factor x^2 + 11x + 18?

Answer 12

Find a factor pair of 18 that sums to 11

x^2 + 11x + 18 = (x + 9)(x + 2)

Question 13

How do you factor x^2 – 8x + 12?

Answer 13

Find a factor pair of 12 that sums to 8, then make both numbers negative

x^2 – 8x + 12 = (x – 6)(x – 2)

Question 14

How do you factor x^2 + 6x – 16?

Answer 14

Find a factor pair of 16 that differs by 6, then make the smaller number negative so that the sum is 6 and the product is -16

x^2 + 6x – 16 = (x + 8)(x – 2)

Question 15

How do you factor x^2 – 5x – 14?

Answer 15

Find a factor pair of 14 that differs by 5, then make the bigger number negative so that the sum is -5 and the product is -14

x^2 – 5x – 14 = (x – 7)(x + 2)

Question 16

How do you factor -2x^2 + 16x – 24?

Answer 16

Factor out the -2 from all terms first, then factor the quadratic expression normally

-2x^2 + 16x – 24 = -2(x^2 -8x + 12) = -2(x – 6)(x – 2)

Question 17

How do you solve a quadratic equation? x^2 + x = 6

Answer 17

(1) Rearrange the equation to make one side equal to 0. The other side will contain a quadratic expression
(2) Factor the quadratic expression. The equation will look like this: (Something)(Something else) = 0…Note: NEVER factor before you set one side equal to 0
(3) Set each factor equal to 0
Something = 0
Something else = 0

Solve for x; each equation will give you a possible solution for the original equation. However, each solution cannot be true at the same time. The variable has multiple possible values. The solutions of a quadratic equation are also called its roots

Example:
x^2 + x = 6
x^2 + x – 6 = 0
(x + 3)(x – 2) = 0

x + 3 = 0
x = -3
OR
x – 2 = 0
x = 2

Question 18

What do you do if an additional condition is placed on the variable in a quadratic equation?
e.g. if y < 0 and y^2 = y + 30, what is the value of y

Answer 18

y^2 – y – 30 = 0
(y – 6)(y + 5) = 0

y – 6 = 0
y = 6

y + 5 = 0
y = -5

The solution must be y = 6

Question 19

How do you solve a quadratic equation with no x term? e.g. x^2 = 25

Answer 19

Take the positive and negative square roots of both sides
x^2 = 25
x = Sqrt(25)
x = 5
x = -5

Question 20

How do you solve a quadratic equation with squared parentheses? e.g. (y + 1)^2 = 16

Answer 20

(y + 1)^2 = 16
(y + 1) = Sqrt(16)
y = 3
y = -5

Question 21

How do you solve an equation with higher powers? e.g. x^3 = 3x^2 – 2x

Answer 21

Look for solutions as if the equation were a typical quadratic: set one side equal to zero, factor as much as you can, then set factors equal to zero

Example:
x^3 – 3x^2 + 2x = 0
x(x^2 – 3x + 2) = 0
x(x – 2)(x – 1) = 0
x = 0
x = 2
x = 1

*Note: NEVER divide an equation by x unless you know for sure that x is not zero. You could be dividing by 0 without realizing it

Question 22

If x does not equal -1, then (x^2 – 2x – 3) / (x + 1) equals what?

Answer 22

(x – 3)(x +1) / (x + 1) = x – 3

Question 23

If x does not equal y, then (y – x)/(x – y) equals what?

Answer 23

-(x – y)/(x – y) = -1

NOTE: Expressions that differ by a sign change are only different by a factor of -1

Question 24

What are the three special products?

Answer 24

MEMORIZE BOTH WAYS!

Square of a Sum
(x + y) = x^2 + 2xy + y2

Square of a Difference
(x – y)^2 = x^2 – 2xy + y^2

Difference of Squares (MOST IMPORTANT)
(x + y)(x – y) = x^2 – y^2

Note: the GMAT likes “square of a sum” and “square of a difference” because they can create quadratic equations that have only one solution

Question 25

What should you do if you see a special product? e.g. 4w^2 – 25z^2

Answer 25

Consider converting it to its other form, especially when you see the difference of squares
(2w + 5z^2)(2w – 5z^2)

Question 26

What should you do if you see something close to special product? n^2 + m^2 = 2nm

Answer 26

Rearrange the equation to try to fit the special product template

n^2 – 2nm + m^2 = 0
(n – m)^2 = 0
n = m

Question 27

What is the quadratic formula?

Answer 27

For any quadratic equation of the form ax^2 + bx + c = 0, where a, b, and c, are constants, the solutions for x are given by the following equation.

x = [-b +/- Sqrt(b^2 – 4ac)] / 2a

Question 28

What is the discriminant?

Answer 28

SQRT(b^2 – 4ac). A portion of the quadratic formula that indicates the number of solutions to the given quadratic equation.

• Positive: two solutions. If the number under the square root sign is positive, there are two solutions to the equation
• Zero: one solution. If the number under the square root sign is zero, there is one solution
• Negative: no solutions. If the number under the square root sign is negative, there are zero solutions

Again, the quadratic formula, and thus the discriminant, is seldom tested on the GMAT.

Question 29

Ch 7, Q 6. Solve the following equation: z^2 = -5z

Answer 29

z^2 – 5z = 0
z(z – 5) = 0
z = 0 OR z = 5

Question 30

Ch 7, Q 18. Simplify the following expression (5x^3 – 5x^2) / (x -1)

Answer 30

(5x^3 – 5x^2) / (x –1)
= 5x^2(x – 1) / (x – 1)
= 5x^2

Question 31

Ch 7, Q 19. Simplify the following expression [(2t – 1) + (2t – 1)^2] / (2t – l)

Answer 31

[(2t – 1) + (2t – 1)^2] / (2t – l)
= (2t – 1)/(2t – 1) + (2t – 1)^2 / (2t – 1)
= 1 + 2t – 1
=2t

Question 32

Ch 7, Q 23. Simplify the following expression (x/5) * (x^5 – x^3)/(x^3 – x^2)

Answer 32

(x/5) * (x^5 – x^3)/(x^3 – x^2)
= (x/5) * x^3(x^2 – 1)/x^2(x – 1)
= (x/5) * x(x – 1)(x + 1)/(x – 1)
= (x/5) * x(x + 1)
= (x/5) * (x^2 + x)
= (x^3 + x^2) / 5