Multiple choice

The number of real roots of the equation $|x|^{6/5}-26|x|^{3/5}-27=0$ is

  1. $0$
  2. $1$
  3. $2$
  4. $3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let y = |x|^(3/5). The equation becomes y^2 - 26y - 27 = 0. Factoring gives (y - 27)(y + 1) = 0. Since y = |x|^(3/5) must be non-negative, y = 27. Then |x|^(3/5) = 27, so |x| = 27^(5/3) = 3^5 = 243. Thus x = 243 or x = -243, giving 2 real roots.

AI explanation

We substitute y = |x|^(3/5) into the equation to obtain a quadratic in y, which is y^2 - 26y - 27 = 0. Factoring this quadratic gives (y - 27)(y + 1) = 0, so the roots are y = 27 and y = -1. Since y = |x|^(3/5) represents a non-negative value for all real x, we discard y = -1 and keep y = 27. Solving |x|^(3/5) = 27 yields |x| = 243, which results in exactly two real roots for x, specifically 243 and -243.

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