Multiple choice

The number of real roots of the equation $\displaystyle \left ( x+3 \right )^{4}+\left ( x+5 \right )^{4}=16$, is:

  1. $0$
  2. $2$
  3. $4$

  4. none
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let y = x + 4. Then the equation becomes (y-1)^4 + (y+1)^4 = 16. Expanding gives 2y^4 + 12y^2 + 2 = 16, or y^4 + 6y^2 - 7 = 0. Letting u = y^2, u^2 + 6u - 7 = 0, so (u+7)(u-1) = 0. Since u = y^2 >= 0, u = 1, so y^2 = 1, y = +/- 1. Thus x+4 = 1 or x+4 = -1, giving x = -3 or x = -5.

AI explanation

Let y = x + 4 to center the terms, transforming the equation into (y - 1)^4 + (y + 1)^4 = 16. Expanding both sides using binomial expansion gives 2(y^4 + 6y^2 + 1) = 16, which simplifies to y^4 + 6y^2 - 7 = 0. Factoring this quadratic in y^2 gives (y^2 + 7)(y^2 - 1) = 0, yielding real values y = 1 and y = -1. Solving for x gives x = -3 and x = -5, confirming there are exactly 2 real roots.

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