Solve each equation in Exercises 41–60 by making an appropriate substitution. (y−8y)2+5(y−8y)−14=0\left( y - \frac{8}{y} \$$right)^2 + 5$$ \left( y - \frac{8}{y} \right) - 14 = 0

Ch. 1 - Equations and Inequalities

Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook

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Blitzer 8th Edition

Ch. 1 - Equations and Inequalities

Problem 59

Chapter 2, Problem 59

Solve each equation in Exercises 41–60 by making an appropriate substitution. ${(y - \frac{8}{y})}^{2} + 5 (y - \frac{8}{y}) - 14 = 0$

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Identify the substitution by letting $ t = y - \frac{8}{y} $. This simplifies the given equation into a quadratic in terms of $ t $.

Rewrite the original equation using the substitution: $ t^2 + 5t - 14 = 0 $.

Solve the quadratic equation $ t^2 + 5t - 14 = 0 $ using the quadratic formula $ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $, where $ a=1 $, $ b=5 $, and $ c=-14 $.

After finding the values of $ t $, substitute back $ t = y - \frac{8}{y} $ to form equations in terms of $ y $: $ y - \frac{8}{y} = t $.

Multiply both sides of each equation by $ y $ to clear the denominator, resulting in quadratic equations in $ y $. Solve these quadratics to find the values of $ y $.

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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, the expression (y - 8/y) can be substituted with a new variable, reducing the equation to a quadratic form that is easier to solve.

Quadratic Equations

A quadratic equation is a polynomial equation of degree two, typically in the form ax² + bx + c = 0. After substitution, the given equation becomes quadratic, allowing the use of factoring, completing the square, or the quadratic formula to find solutions.

Solving Rational Expressions

The original expression involves a rational term (8/y), so after finding solutions for the substituted variable, it is important to back-substitute and solve for y. This may involve solving rational equations and checking for extraneous solutions where the denominator is zero.

Solve each equation in Exercises 41–60 by making an appropriate substitution. (y−8y)2+5(y−8y)−14=0\(\left\)( y - \(\frac{8}{y}\) \(\right\))^2 + 5 \(\left\)( y - \(\frac{8}{y}\) \(\right\)) - 14 = 0