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Ch. 1 - Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Non è quello che usi tu?Cambia libro di testo
Capitolo 2, Problema 43

Solve each equation in Exercises 41–60 by making an appropriate substitution. 9x4=25x2−169x^4 = 25x^2 - 16

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1
Identify the substitution to simplify the equation. Notice that the equation involves terms with \(x^4\) and \(x^2\). Let \(u = x^2\), so that \(x^4 = (x^2)^2 = u^2\).
Rewrite the original equation \(9x^4 = 25x^2 - 16\) in terms of \(u\): it becomes \(9u^2 = 25u - 16\).
Bring all terms to one side to set the equation equal to zero: \(9u^2 - 25u + 16 = 0\).
Solve the quadratic equation \(9u^2 - 25u + 16 = 0\) using the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=9\), \(b=-25\), and \(c=16\).
After finding the values of \(u\), substitute back \(u = x^2\) and solve each resulting equation \(x^2 = u\) for \(x\) by taking the square root, remembering to consider both positive and negative roots.

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