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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 52

Solve each equation in Exercises 41–60 by making an appropriate substitution. x25+x15−6=0x^{\(\frac{2}{5}\)} + x^{\(\frac{1}{5}\)} - 6 = 0

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1
Identify the substitution to simplify the equation. Notice that the exponents are fractional and related: \(x^{2/5}\) and \(x^{1/5}\). Let \(u = x^{1/5}\), so that \(u^2 = x^{2/5}\).
Rewrite the original equation in terms of \(u\): replace \(x^{2/5}\) with \(u^2\) and \(x^{1/5}\) with \(u\). The equation becomes \(u^2 + u - 6 = 0\).
Solve the quadratic equation \(u^2 + u - 6 = 0\) using factoring, completing the square, or the quadratic formula. This will give you the possible values of \(u\).
After finding the values of \(u\), recall that \(u = x^{1/5}\). To find \(x\), raise both sides of the equation \(u = x^{1/5}\) to the 5th power, resulting in \(x = u^5\).
Calculate \(x\) for each value of \(u\) found in step 3 by computing \(x = u^5\). These are the solutions to the original equation.

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Substitution Method

Substitution involves replacing a complex expression with a simpler variable to transform the equation into a more familiar form, often a polynomial. This technique simplifies solving equations that contain complicated terms, such as fractional powers, by reducing them to quadratic or linear equations.
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