The number of real roots of the equation x6 - 7x3 - 8 = 0 is
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1
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2
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3
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6
Reveal answer
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B
Correct answer
Explanation
Let y = x^3. The equation becomes y^2 - 7y - 8 = 0. Factoring gives (y - 8)(y + 1) = 0, so y = 8 or y = -1. This means x^3 = 8 (x = 2) or x^3 = -1 (x = -1). Both are real roots, so there are 2 real roots.
AI explanation
We can treat x to the sixth power minus 7x cubed minus 8 = 0 as a quadratic in terms of y = x cubed, rewriting it as y squared - 7y - 8 = 0. Factoring this quadratic gives (y - 8)(y + 1) = 0, so the values for y are 8 and -1. Substituting x cubed back in gives x cubed = 8, which yields the real root x = 2, and x cubed = -1, which yields the real root x = -1. Since these are the only real values that satisfy the equation, the number of real roots is 2.