Solving Nonlinear Equations by Substitution
Example
Solve for w: w -2 - 2w -1 = 15
Solution Step 1 Write the equation in quadratic form.
Subtract 15 from both sides. Write w -2 as (w -1)2.
Step 2 Use an appropriate “u†substitution. Substitute u for w
-1.
Step 3 Solve the resulting equation.
Factor the new equation.
Use the Zero Product Property. Solve each equation for u.
Step 4 Substitute the original
expression for u.
Step 5 Solve for the original variable. |
w -2 - 2w -1
w -2 - 2w -1 - 15
(w -1)2 - 2(w -1)1 - 15
u2 - 2u - 15
(u - 5)(u + 3)
u - 5 = 0 or u + 3
u = 5 or u
w -1 = 5 or w
-1
|
= 15
= 0
= 0
= 0
= 0
= 0
= -3 = -3 |
Use the definition of a negative exponent.
|
|
= -3 |
Multiply by the LCD, w. |
1 = 5w or 1 |
= -3w |
Divide by the coefficient of w. |
|
= w |
So, there are two solutions:
The equation w -2 - 2w -1 = 15 written
in standard form is w -2 - 2w -1 - 15 = 0. The graph of
the corresponding function, f(w) = w -2 - 2w -1 - 15
is shown.
The graph crosses the w-axis at two locations,
Note that w -1 can be written as
Since w is in the denominator it cannot equal 0. Therefore the line w = 0 is a
vertical asymptote.
|