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Ch. 1 - Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
Chapter 2, Problem 55

Solve each equation in Exercises 41–60 by making an appropriate substitution. (x - 5)2 - 4(x - 5) - 21 = 0

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Identify the substitution to simplify the equation. Let \( u = x - 5 \). This transforms the equation into a quadratic in terms of \( u \).
Rewrite the original equation using the substitution: \( (x - 5)^2 - 4(x - 5) - 21 = 0 \) becomes \( u^2 - 4u - 21 = 0 \).
Solve the quadratic equation \( u^2 - 4u - 21 = 0 \) by factoring, completing the square, or using the quadratic formula.
After finding the values of \( u \), substitute back \( u = x - 5 \) to get equations in terms of \( x \).
Solve each resulting linear equation for \( x \) to find the solutions to the original equation.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Substitution Method

The substitution method involves replacing a complex expression with a single variable to simplify the equation. In this problem, letting y = (x - 5) transforms the equation into a quadratic in terms of y, making it easier to solve.
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Quadratic Equations

A quadratic equation is a second-degree polynomial equation of the form ax² + bx + c = 0. Recognizing the transformed equation as quadratic allows the use of factoring, completing the square, or the quadratic formula to find solutions.
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Back-Substitution

After solving the quadratic equation for the substituted variable, back-substitution involves replacing the variable with the original expression to find the values of x. This step ensures the solutions correspond to the original equation.
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