I. Completing the Square:

if x

^{2}+ bx = c, thenx

^{2}+ bx + (b/2)^{2}= c + (b/2)^{2}so:

(x + b/2)

^{2}= c + (b^{2})/4Example:

x

^{2}+ 4x - 32 = 0x

^{2}+ 4x = 32therefore

*b*= 4 so:x

^{2}+ 4x + (4/2)^{2}= 32 + (4/2)^{2}(x + 2)

^{2}= 32 + 4(x + 2)

^{2}= 36take the square root of both sides:

x + 2 = ± √36

x + 2 = ±6

x = -2 ±6

so x = 4 and x = -8

Now check your answers:

(-8)

^{2}+ 4(-8) - 32 = 064 - 32 - 32 = 0

0 = 0

(4)

^{2}+ 4(4) - 32 = 016 + 16 - 32 = 0

0 = 0

They both work so our answers are x = 4 and x = -8

Example 2:

x

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

^{2}- 2x = 3x

^{2}- 2x + (-2/2)^{2}= 3 + (-2/2)^{2}x

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

^{2}= 4x - 1 = ±√4

x = 1 ±2

x = 1 + 2 = 3

x = 1 - 2 = -1

check answers:

(3)

^{2}- 2(3) - 3 = 09 - 6 - 3 = 0

0 = 0

(-1)

^{2}- 2(-1) - 3 = 01 + 2 - 3 = 0

0 = 0

II. Quadratic Formula:

if ax

^{2}+ bx + c = 0 thenExample:

9x

^{2}- 12x - 14 = 0x = (12 ± √(144 - 4(9)(-18)))/(2(9))

x = (12 ± √648)/18

x = (12 ± √324 √2)/18

x = (12 ± 18 √2)/18

x = (2 ± 3 √2)/3

x = 2/3 ± √2

III. Solving an Equation of Quadratic Type:

x

^{4}+ 2x^{3}- 8x - 16 = 0 by groupingx

^{3}(x + 2) - 8(x + 2) = 0(x

^{3}- 8 ) ( x + 2) = 0x

^{3}= 8x = 2

x + 2 = 0

x = -2

now check the answers:

they both check!

IV. By factoring more difficult equations:

4x

^{2}(x - 1)^{1/3}+ 6x (x - 1)^{4/3}= 02x [2x (x - 1)

^{1/3}+ 3 (x - 1)^{4/3}] = 0remember that (x - 1)

^{4/3}= (x - 1)^{3/3}(x - 1)^{ 1/3}and 3/3 = 1 so:

2x(x - 1)

^{1/3}[2x + 3 (x - 1)^{3/3}] = 02x(x - 1)

^{1/3}[ 2x + 3x - 3] = 02x(x - 1)

^{1/3}[5x - 3] = 0so:

2x = 0

x = 0

(x - 1)

^{1/3}= 0x - 1 = 0

x = 1

5x - 3 = 0

5x = 3

x = 3/5

check your answers!

V. Solving an equation involving a radical:

√(-5x + 4) - x = 6

√(-6x + 4) = x + 6

now square both sides:

(√(-6x + 4))

^{2}= (x + 6)^{2}-6x + 4 = x

^{2}+ 12x + 360 = x

^{2}+ 18 x + 320 = (x + 16)(x + 2)

x = -16 and x = -2

check both answers:

√(-6 (-16) + 4) - (-16) = 6

√(96 + 4 ) + 16 = 6

10 + 16 = 6

not true

√(-6(-2) + 4) - (-2) = 6

√(12 + 4) + 2 = 6

√16 + 2 = 6

4 + 2 = 6

this one works! so

x = -2 is the only answer!

VI. Solving an Equation Involving Two radicals:

√x + √(x - 20) = 10

√ x = 10 - √(x - 20)

(√x)

^{2}= (10 - √(x - 20))^{2}x = 100 - 20√(x - 20) + (x - 20)

0 = 80 - 20√(x - 20)

-80 = -20√(x - 20)

4 = √(x - 20)

4

^{2}= x - 2016 = x - 20

36 = x

Now check your answer.

√36 + √(36 - 20) = 10

6 + 4 = 10 so it checks

VII. Solving an Equation with Rational Exponents

Example:

(x - 5)

^{2/3}= 16to undo exponents, multiply by the reciprocal

((x - 5)

^{2/3})^{3/2}= 16^{3/2}x - 5 = 64

x = 69

Check your answer.

(69 - 5)

^{2/3}= 1664

^{2/3}= 1616 = 16 so it checks

VIII. Solving an Equation Involving Absolute Value:

recall: |x| = 9

x = 9 and x = -9 so

|3x + 2 | = 7

3x + 2 = 7 and 3x + 2 = -7

3x = 5 and 3x = -9

x = 5/3 and x = -3