A.5
Ex. 1
Solving Equations
Solve.
6(x - 1) + 4 = 7x + 1
x = -3
Ex. 2
x 3x
+
=2
3 4
Multiply each term
by the LCD -----> 12
x
3x
(12) + (12) = (12)2
3
4
4x + 9x = 24
13x = 24
24
x=
13
Ex. 3
An equation with an Extraneous Solution
1
3
6x
=
! 2
x!2 x+2 x !4
Again, mult. Each
term by the LCD.
1
3
6x
(x + 2)( x ! 2) =
(x + 2)( x ! 2) ! 2 (x + 2)( x ! 2)
x!2
x+2
x !4
x + 2 = 3(x - 2) - 6x
x + 2 = 3x - 6 - 6x
4x = -8
x = -2
But, x = -2 yields a den.
of zero, therefore there
are no solutions.
Ex. 4
A.
Solving a Quadratic by Factoring
6x2 = 3x
6x2 - 3x = 0
3x(2x - 1) = 0
First, take the 3x to the
other side and then factor.
Set both factors = to 0.
3x = 0
x=0
B.
2x - 1 = 0
2x = 1
x = 1/2
9x2 - 6x + 1 = 0
(3x -1)(3x - 1) = 0
1
x=
3
First, factor.
Set = to 0 and solve.
Ex. 5
A.
Solving quadratic equations.
4x2
= 12
x2 = 3
x=± 3
Ex. 6a
B.
(x - 2)2 = 5
x!2= ± 5
x = 2± 5
Completing the Square
First, take 2 to the
x2 - 6x + 2 = 0
other side.
x2 - 6x
= -2
To complete the square
x2 - 6x + 9 = -2 + 9 take half the x-term and
square it. Add it to both
(x - 3)2 = 7
sides.
x !3 = ± 7
x = 3± 7
Ex. 6b
2
Completing the Square when the
leading coefficient is not 1
Divide each term by 3.
3x " 4 x " 5 = 0
4x 5
2
Take 5/3 to the other side.
x "
" =0
3 3
4x 5
2
Now, complete the square.
x "
=
3
3 2
2
4 x # 2%
5 # 2%
Take the square
2
x "
+
= +
root of both sides.
$
&
$
&
3
3
3
3
2
2
19
2%
19 #
2%
19
#
x= ±
x"
=
x"
=±
$
3
3
3&
9 $
3&
9
Ex. 7
Use the Quadratic Formula to solve
x2 + 3x - 9 = 0
2
! b ± b ! 4ac
x=
2a
2
! 3 ± 3 ! 4(1)(! 9 )
x=
2(1)
! 3 ± 45
x=
2
!3±3 5
x=
2
Ex. 8
Solve by factoring.
x4 - 3x2 + 2 = 0
Factor
Set both factors = 0
(x2 - 2)(x2 - 1) = 0
or factor again.
x2 - 2 = 0
x2 - 1 = 0
x2 = 2
x2 = 1
x=± 2
x = ±1
Ex. 9
Solve by grouping.
x3 - 3x2 - 3x + 9 = 0
x2(x - 3) - 3(x - 3) = 0
(x - 3)(x2 - 3) = 0
x=3
x=± 3
Factor out an (x - 3)
Ex. 10
Solving a Radical
2x + 7 ! x = 2
Isolate the radical.
2x + 7 = x + 2
Now square both sides.
2x + 7 = x2 + 4x + 4
0 = x2 + 2x - 3
0 = (x + 3)(x - 1)
Factor or use quad.
formula
Possible answers for x are - 3 and 1.
Check them in the original equation to see if
they work.
Only x = 1 works!
Ex. 11
An equation Involving Two Radicals
2x + 6 ! x + 4 = 1
Isolate the more
complicated rad.
Square both sides.
2x + 6 = x + 4 +1
2 x + 6 = (x + 4 )+ 2 x + 4 + 1
Once again, isolate the radical.
x +1 = 2 x + 4
x2 + 2x + 1 = 4(x + 4)
x2 - 2x - 15 = 0
Square both sides.
(x - 5)(x + 3) = 0
Only x = 5 works.
Ex. 12
Solving an Equation Involving
Absolute Value
2
x ! 3 x = !4 x + 6
x2 - 3x = -4x + 6
x2 + x - 6 = 0
(x + 3)(x - 2) = 0
Split into 2
equations.
x2 - 3x = 4x -6
Now solve
them for x.
x2 - 7x + 6 = 0
(x - 1)(x - 6) = 0
Possible answers are -3, 2, 1, and 6. Which ones work?
-3 and 1