REVIEW OF SOLVING ALGEBRAIC EQUATIONS

```REVIEW OF SOLVING ALGEBRAIC EQUATIONS
Solving by Factoring:
a) x2 + 6x – 16 = 0
Look for two numbers that multiply to be –16 and add to be 6.
–16 and 1
–8 and 2
–4 and 4
–1 and 16
–2 and 8

x2 + 6x – 16 = 0
(x – 2)(x + 8) = 0
So, either x – 2 = 0 or
x = 2 or
x+8 = 0
x = –8
b) 2x2 – x – 3 = 0
This will factor into something of the form: (2x ? )(x ? )
Use trial and error with factors of –3.
(2x – 3)(x + 1)
(2x + 1)(x – 3)
(2x + 3)(x – 1)
(2x – 1)(x + 3)

=
=
=
=
2x2 + 2x – 3x – 3 = 2x2 – x – 3
2x2 – 6x + x – 3 = 2x2 – 5x – 3
2x2 – 2x +3x – 3 = 2x2 + x – 3
2x2 + 6x – x – 3 = 2x2 + 5x – 3
 Yes
 No
 No
 No
2x2 – x – 3 = 0
(2x – 3)(x + 1) = 0
So, either 2x – 3 = 0
x = 3/2
or x + 1 = 0
or
x = –1
c) 9x2 – 6x + 1 = 0
This will factor into something of the form: (9x ? )(x ? )
Use trial and error with factors of 1.
or
(3x ? )(3x ? )
(9x + 1)(x + 1) = 9x2 + 9x + x + 1 = 9x2 + 10x + 1  No
(9x – 1)(x – 1) = 9x2 – 9x – x + 1 = 9x2 – 10x + 1  No
(3x + 1)(3x + 1) = 9x2 + 3x +3x + 1 = 9x2 + 6x + 1  No
(3x – 1)(3x – 3) = 9x2 – 3x – 3x + 1 = 9x2 – 6x + 1  Yes

9x2 – 6x + 1 = 0
(3x – 1)(3x – 1) = 0
So, 3x – 1 = 0
 x = 1/3
d) 4x2 – 64 = 0
4x2 – 64 = 0
4(x2 – 16) = 0
So,
(factor out a 4)
x2 – 16 = 0
(x + 4)(x – 4) = 0 (using the difference of squares formula)
So, either x + 4 = 0
x=–4
or
or
x–4 = 0
x=4
e) 3x2 – 5x = 0
3x2 – 5x = 0
x(3x – 5) = 0
So, either x = 0
x=0
(factor out an x)
or
or
3x – 5 = 0
x = 5 /3
Solving using Square Roots:
a) x2 – 49 = 0
x2 – 49 = 0
x2 = 49
x 2   49
x=&plusmn;7
(square root both sides)
b) 5x2 – 10 = 0
5x2 – 10 = 0
5x2 = 10
x2 = 2
x2   2
x=&plusmn;
2
(square root both sides)
c) (3x + 5)2 = 8
(3x + 5)2 = 8
(3 x  5) 2   8
(square root both sides)
3x  5   8
3x   8  5
x 
 8 5
5  8
5  2 2


3
3
3
Solving by Completing the Square:
Steps: 1. Move c over
2. Take half of b, square it, add that number to both sides
3. Factor and solve
Note: If a ≠ 1, you must divide everything by a first.
a) x2 – 6x – 11 = 0
x2 – 6x – 11 = 0
x2 – 6x + _____ = 11
(move over the 11 and leave a space)
x2 – 6x + __9__ = 11 + 9
(half of 6 is 3. 3 squared is 9. Add 9 to both sides)
x2 – 6x + 9 = 20
(x – 3)(x – 3) = 20
(left side should always factor into “twin” factors)
(x – 3) 2 = 20
x  3   20
x   20  3  3  20
 3 2 5
b) 2x2 + 3x – 4 = 0
2 x 2  3x  4  0
2 2 3
4
x  x  0
2
2
2
2
3
x  2x2  0
(divide everything by 2 first)
x 2  32 x  _____  2 (move the 2 over)
x 2  32 x 
9
16
 2  169
x 2  32 x 
9
16

( x  34 ) 2 
x
3
4
 
x
3
4

x 
 41
4
41
16
41
16
41
16
 41
4
 43 
3  41
4
(half of
3
2
=
1
2
 32  34 .
 34 
2