Question

Solve (x + 5)^(3/2)= (x - 1)^3

Answer 1

Answer:

  x = 4

Step-by-step explanation:

For an equation of this sort, my first step is to rewrite it so the solutions are the x-intercepts of the graph:

  (x +5)^(3/2) -(x +1)^3 = 0

The graph is attached. The one real solution is x=4.

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The solution method for something like this is necessarily ad hoc. Here, it appears that progress can be made by raising both sides of the equation to the 2/3 power:

  ((x +5)^(3/2))^(2/3) = ((x -1)^3)^(2/3)

  x +5 = (x -1)^2

Now, you have an ordinary quadratic that can be solved using any of the usual methods.

  x^2 -2x +1 = x +5 . . . . swap sides, expand the square

  x^2 -3x -4 = 0 . . . . . . put in standard form

  (x -4)(x +1) = 0 . . . . . . factor

The solutions to this are ...

  x = 4, x = -1 . . . . . . . values of x that make the factors zero

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To see if either of these solutions is extraneous, we can check them:

  (4 +5)^(3/2) = (4 -1)^3 . . . . . substitute 4 for x

  9^(3/2) = 3^3 . . . . . . . . . true

  (-1 +5)^(3/2) = (-1 -1)^3 . . . . . substitute -1 for x

  4^(3/2) = (-2)^3

  8 = -8 . . . . . . . . . . . false; x = -1 is extraneous

The solution is x = 4.

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Additional comment

You can eliminate the x=-1 "solution" by considering the domains of the left- and right-side expressions. The 3/2 power is equivalent to the square root of the cube. That will generally only be defined for non-negative values (x ≥ -5). The cube on the right will only be positive for x > 1, so the practical domain of this equation is x ≥ 1.