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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 144

Solve for x: x^(5/6) + x^(2/3) - 2x^(1/2) = 0

Guida verificata passo dopo passo
1
Rewrite the exponents in terms of a common base. Notice that all the exponents are fractions with a denominator of 6. Rewrite the terms as follows: \(x^{5/6} = (x^{1/6})^5\), \(x^{2/3} = (x^{1/6})^4\), and \(x^{1/2} = (x^{1/6})^3\).
Let \(y = x^{1/6}\). This substitution simplifies the equation to \(y^5 + y^4 - 2y^3 = 0\).
Factor the simplified equation. Start by factoring out the greatest common factor, \(y^3\), which gives \(y^3(y^2 + y - 2) = 0\).
Solve for \(y\). The equation \(y^3 = 0\) gives \(y = 0\). For the quadratic \(y^2 + y - 2 = 0\), factor it as \((y + 2)(y - 1) = 0\), which gives \(y = -2\) and \(y = 1\).
Substitute back \(y = x^{1/6}\) into each solution. For \(y = 0\), \(x^{1/6} = 0\), so \(x = 0\). For \(y = 1\), \(x^{1/6} = 1\), so \(x = 1\). Discard \(y = -2\) because \(x^{1/6}\) cannot be negative for real numbers. The solutions are \(x = 0\) and \(x = 1\).

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Exponents and Fractional Powers

Understanding exponents, particularly fractional powers, is crucial in this equation. The expression x^(5/6) indicates that x is raised to the power of 5/6, which can be interpreted as the sixth root of x raised to the fifth power. This concept is essential for manipulating and simplifying the equation effectively.
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Factoring Polynomials

Factoring is a key technique in solving polynomial equations. In this case, the equation can be rearranged and factored to find the values of x that satisfy the equation. Recognizing common factors or using substitution can simplify the process of finding solutions.
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Roots of Equations

Finding the roots of an equation involves determining the values of x that make the equation equal to zero. This concept is fundamental in algebra, as it allows us to identify solutions to polynomial equations. Techniques such as the Rational Root Theorem or numerical methods may be employed to find these roots.
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