## Monday, September 10, 2007

### Precalculus P4.b Completing the square and

I. Completing the Square:

if x2 + bx = c, then

x2 + bx + (b/2)2 = c + (b/2)2

so:

(x + b/2)2 = c + (b2)/4

Example:

x2 + 4x - 32 = 0

x2 + 4x = 32

therefore b = 4 so:

x2 + 4x + (4/2)2 = 32 + (4/2)2

(x + 2)2 = 32 + 4

(x + 2)2 = 36

take the square root of both sides:

x + 2 = ± √36

x + 2 = ±6

x = -2 ±6

so x = 4 and x = -8

(-8)2 + 4(-8) - 32 = 0

64 - 32 - 32 = 0

0 = 0

(4)2 + 4(4) - 32 = 0

16 + 16 - 32 = 0

0 = 0

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

Example 2:

x2 - 2x - 3 = 0

x2 - 2x = 3

x2 - 2x + (-2/2)2 = 3 + (-2/2)2

x2 - 2x + (-1)2 = 3 + 1

(x - 1)2 = 4

x - 1 = ±√4

x = 1 ±2

x = 1 + 2 = 3

x = 1 - 2 = -1

(3)2 - 2(3) - 3 = 0

9 - 6 - 3 = 0

0 = 0

(-1)2 - 2(-1) - 3 = 0

1 + 2 - 3 = 0

0 = 0

if ax2 + bx + c = 0 then

Example:
9x2 - 12x - 14 = 0
x = (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:
x4 + 2x3 - 8x - 16 = 0 by grouping
x3 (x + 2) - 8(x + 2) = 0
(x3 - 8 ) ( x + 2) = 0
x3 = 8
x = 2
x + 2 = 0
x = -2
they both check!
IV. By factoring more difficult equations:
4x2(x - 1)1/3 + 6x (x - 1)4/3 = 0
2x [2x (x - 1)1/3 + 3 (x - 1)4/3] = 0
remember 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 ] = 0
2x(x - 1)1/3 [ 2x + 3x - 3] = 0
2x(x - 1)1/3 [5x - 3] = 0
so:
2x = 0
x = 0
(x - 1)1/3 = 0
x - 1 = 0
x = 1
5x - 3 = 0
5x = 3
x = 3/5
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 = x2 + 12x + 36
0 = x2 + 18 x + 32
0 = (x + 16)(x + 2)
x = -16 and x = -2
√(-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)
42 = x - 20
16 = x - 20
36 = x
√36 + √(36 - 20) = 10
6 + 4 = 10 so it checks

VII. Solving an Equation with Rational Exponents
Example:
(x - 5)2/3 = 16
to undo exponents, multiply by the reciprocal
((x - 5)2/3)3/2 = 163/2
x - 5 = 64
x = 69
(69 - 5)2/3 = 16
642/3 = 16
16 = 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