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Difficulty: Very hardAlgebraic Exponents and Radicals

Which of the following values of xx are solutions to the equation (x−2)43−5(x−2)23+4=0(x - 2)^{\frac{4}{3}} - 5(x - 2)^{\frac{2}{3}} + 4 = 0? Select all that apply.

  1. -6Answer
  2. 1Answer
  3. 10Answer
  4. D
    -2
  5. E
    6

Answer

The correct values of xx that satisfy the equation are −6-6, 11, and 1010.
Substituting u=(x−2)23u = (x - 2)^{\frac{2}{3}} yields u2−5u+4=0u^2 - 5u + 4 = 0, which factors as (u−1)(u−4)=0(u - 1)(u - 4) = 0, giving u=1u = 1 and u=4u = 4. Solving (x−2)23=1(x - 2)^{\frac{2}{3}} = 1 gives (x−2)2=1  ⟹  x−2=±1(x - 2)^2 = 1 \implies x - 2 = \pm 1, yielding x=3x = 3 and x=1x = 1. Solving (x−2)23=4(x - 2)^{\frac{2}{3}} = 4 gives (x−2)2=64  ⟹  x−2=±8(x - 2)^2 = 64 \implies x - 2 = \pm 8, yielding x=10x = 10 and x=−6x = -6. Thus, the values −6-6, 11, and 1010 are all valid solutions.

Step-by-Step Solution

1
Perform a substitution to rewrite the equation in quadratic form.
Let u=(x−2)23u = (x - 2)^{\frac{2}{3}}. Then u2=(x−2)43u^2 = (x - 2)^{\frac{4}{3}}, giving u2−5u+4=0u^2 - 5u + 4 = 0.
Recognizing quadratic structure simplifies equations with rational exponents.
2
Solve the quadratic equation for uu.
(u−1)(u−4)=0  ⟹  u=1(u - 1)(u - 4) = 0 \implies u = 1 or u=4u = 4.
Factoring determines the values of the substituted variable uu.
3
Solve for xx when u=1u = 1.
(x−2)23=1  ⟹  (x−2)2=13=1  ⟹  x−2=±1  ⟹  x=3(x - 2)^{\frac{2}{3}} = 1 \implies (x - 2)^2 = 1^3 = 1 \implies x - 2 = \pm 1 \implies x = 3 or x=1x = 1.
Raising both sides to the power of 32\frac{3}{2} requires taking both positive and negative roots because the numerator of the power is even.
4
Solve for xx when u=4u = 4.
(x−2)23=4  ⟹  (x−2)2=43=64  ⟹  x−2=±8  ⟹  x=10(x - 2)^{\frac{2}{3}} = 4 \implies (x - 2)^2 = 4^3 = 64 \implies x - 2 = \pm 8 \implies x = 10 or x=−6x = -6.
Squaring and taking square roots yields two solutions, 1010 and −6-6.
5
Match calculated solutions with the given choices.
The solutions present among the options are −6-6, 11, and 1010.
Comparing all valid algebraic solutions to the available choices identifies all correct options.

Key Concept

Solving quadratic-form equations with fractional exponents and accounting for negative base branches when taking even roots.
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Topics

Algebraic Exponents and Radicals

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