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Algebra I

Solving Quadratic Equations

Lesson

A quadratic equation has an x2x^2 term and is set equal to zero:

ax2+bx+c=0ax^2 + bx + c = 0

It usually has two solutions. There are two methods to know.

Method 1: Solve by factoring

Factor the polynomial, then use the zero-product property: if a product equals zero, at least one factor must be zero.

Worked example 1

x2+5x+6=0x^2 + 5x + 6 = 0

Factor:

(x+2)(x+3)=0(x + 2)(x + 3) = 0

Set each factor to zero:

x+2=0orx+3=0x + 2 = 0 \quad \text{or} \quad x + 3 = 0
x=−2orx=−3x = -2 \quad \text{or} \quad x = -3

Method 2: Solve by square roots

When the equation has the form x2=kx^2 = k, take the square root of both sides. Don’t forget the ± — both the positive and negative roots are solutions.

Worked example 2

x2=49x^2 = 49
x=±7x = \pm 7

So x=7x = 7 or x=−7x = -7.

How to type your answer

Enter the two solutions separated by a comma. Order doesn’t matter. Examples: -2,-3, 7,-7, 5,-2. If both solutions are the same (a double root), enter the number twice: 3,3.

Practice

Work through these. Stuck? Click Get a hint.

Warm-Up

Quick problems to get going.

Problem 1

x2=9x^2 = 9

Problem 2

x2=25x^2 = 25

Problem 3

x2=100x^2 = 100

Problem 4

(x−4)(x+2)=0(x - 4)(x + 2) = 0

Practice

Standard problems matching the lesson.

Problem 5

x2=16x^2 = 16

Problem 6

x2=36x^2 = 36

Problem 7

x2=64x^2 = 64

Problem 8

(x+3)(x−5)=0(x + 3)(x - 5) = 0

Problem 9

x2+5x+6=0x^2 + 5x + 6 = 0

Problem 10

x2+7x+10=0x^2 + 7x + 10 = 0

Problem 11

x2−7x+12=0x^2 - 7x + 12 = 0

Problem 12

x2+2x−8=0x^2 + 2x - 8 = 0

Problem 13

x2−x−12=0x^2 - x - 12 = 0

Problem 14

x2−9=0x^2 - 9 = 0

Problem 15

x2−49=0x^2 - 49 = 0

Problem 16

x2−8x+15=0x^2 - 8x + 15 = 0

Challenge

Harder problems — edge cases, trickier numbers, multiple steps.

Problem 17

x2+4x−21=0x^2 + 4x - 21 = 0

Problem 18

x2−11x+28=0x^2 - 11x + 28 = 0

Problem 19

x2+6x+9=0x^2 + 6x + 9 = 0

Problem 20

x2−10x+25=0x^2 - 10x + 25 = 0

Problem 21

x2−144=0x^2 - 144 = 0

Problem 22

x2+3x−40=0x^2 + 3x - 40 = 0

Practice

Standard problems matching the lesson.

Problem 23

A ball's height (m) at second t is h = -t^2 + t + 6. Solve -t^2 + t + 6 = 0 for both values of t.

Problem 24

A rectangle's length is x and width is (x - 1). Area is 12 sq ft. Solve x(x - 1) = 12 for both values of x.

Challenge

Harder problems — edge cases, trickier numbers, multiple steps.

Problem 25

A box's two dimensions satisfy x(x + 1) = 30. Solve for both values of x.

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