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Solving Quadratic Equations by Completing the Square
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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.

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