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 â€œuâ€ 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
u^{2}  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 waxis 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.
