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Precalculus Review
Copyright © Cengage Learning. All rights reserved.

2. Copyright © Cengage Learning. All rights reserved.
0.6
Solving Miscellaneous Equations
Copyright © Cengage Learning. All rights reserved.

3. Solving Miscellaneous Equations
Solving an Equation of the Form P  Q = 0
If a product is equal to 0, then at least one of the factors must be 0. That is, if P  Q = 0, then either P = 0 or Q = 0.
Quick Example
x5 – 4x3 = 0
x3(x2 – 4) = 0
Either x3 = 0 or x2 – 4 = 0
x = 0, 2 or –2.
Factor the left-hand side.
Either P = 0 or Q = 0.
Solve the individual equations.

4. Example 1 – Solving by Factoring
Solve 12x(x2 – 4)5(x2 + 2)6 + 12x(x2 – 4)6(x2 + 2)5 = 0.
Solution:
Again, we start by factoring the left-hand side:
12x(x2 – 4)5(x2 + 2)6 + 12x(x2 – 4)6(x2 + 2)5
= 12x(x2 – 4)5(x2 + 2)5[(x2 + 2) + (x2 – 4)]
= 12x(x2 – 4)5(x2 + 2)5(2x2 – 2)
= 24x(x2 – 4)5(x2 + 2)5(x2 – 1).

5. Example 1 – Solution Setting this equal to 0, we get:
cont’d
Setting this equal to 0, we get:
24x(x2 – 4)5(x2 + 2)5(x2 – 1) = 0,
which means that at least one of the factors of this product must be zero.
Now it certainly cannot be the 24, but it could be the x:
x = 0 is one solution.
It could also be that
(x2 – 4)5 = 0 or
x2 – 4 = 0,
which has solutions x = ±2.

6. Example 1 – Solution
cont’d
Could it be that (x2 + 2)5 = 0? If so, then x2 + 2 = 0, but this
is impossible because x2 + 2 ≥ 2, no matter what x is.
Finally, it could be that x2 – 1 = 0, which has solutions x = ±1.
This gives us five solutions to the original equation:
x = –2, –1, 0, 1, or 2.

7. Solving Miscellaneous Equations
Solving an Equation of the Form P/Q = 0
If = 0, then P = 0.
How else could a fraction equal 0? If that is not convincing, multiply both sides by Q (which cannot be 0 if the quotient is defined).

8. Solving Miscellaneous Equations
Quick Example
(x + 1)(x + 2)2 – (x + 1)2(x + 2) = 0
(x + 1)(x + 2)[(x + 2) – (x + 1)] = 0
(x + 1)(x + 2)(1) = 0
Either x + 1 = 0 or x + 2 = 0,
x = –1 or x = –2
x = –1
If = 0, then P = 0.
Factor.
x = –2 does not make sense in the original equation: it makes the denominator 0. So it is not a solution and
x = –1 is the only solution.