If (-2, y) lies on the graph of y = 4x, then y = -16 1/16 16
2 answers:
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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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<em>Additional comment</em>
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.
