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Ch. 1 - Equations and Inequalities
Lial - College Algebra 13th Edition
Lial13th EditionCollege AlgebraISBN: 9780136881063Not the one you use?Change textbook
Chapter 2, Problem 97

Solve each equation. 6(x+2)4-11(x+2)2=-4

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1
Start by making a substitution to simplify the equation. Let \(y = (x+2)^2\). This transforms the original equation \$6(x+2)^4 - 11(x+2)^2 = -4\( into an equation in terms of \)y\(: \)6y^2 - 11y = -4$.
Rewrite the equation in standard quadratic form by moving all terms to one side: \$6y^2 - 11y + 4 = 0$.
Use the quadratic formula to solve for \(y\). Recall the quadratic formula is \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=6\), \(b=-11\), and \(c=4\).
After finding the values of \(y\), substitute back \(y = (x+2)^2\) to get equations of the form \((x+2)^2 = y_1\) and \((x+2)^2 = y_2\).
Solve each equation for \(x\) by taking the square root of both sides, remembering to consider both the positive and negative roots: \(x + 2 = \pm \sqrt{y_i}\). Then isolate \(x\) by subtracting 2.

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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 + 2)^2 reduces the quartic equation to a quadratic form, making it easier to solve.
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Solving Quadratic Equations

Once the equation is rewritten as a quadratic in terms of y, techniques such as factoring, completing the square, or using the quadratic formula can be applied to find the values of y. These solutions are then used to find x.
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Back-Substitution and Solving for x

After finding the values of y, substitute back y = (x + 2)^2 to solve for x. This involves taking square roots and considering both positive and negative roots, leading to the final solutions for x.
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