Answer:
- x = -4 . . . actual solution
- x = -7 . . . extraneous solution
Explanation:
I like to solve these using a graphing calculator. You can plot the left side and right side (as below) and look for points of intersection. Here, the point of intersection is at x = -4, so that is the only solution.
If we plot the negative branch of the square root function, we can see that an extraneous solution would be x = -7. We expect to get that solution as part of a conventional solution process for this equation.
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My preference for solving these graphically is to write the equation as ...
(something) = 0
A graphing calculator is usually very good at finding zeros of a function to high precision, so that works well. Here, I might use ...
f(x) = √(x +8) -6 -x
and look for x that makes f(x) = 0. (see second attachment)
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The usual way to solve these is to isolate the radical, then raise the equation to a power that eliminates the radical. Here, we can add 6 to get ...
√(x +8) = x +6
x +8 = (x +6)² . . . . . . . .square both sides of the equation
x² +11x +28 = 0 . . . . . . eliminate parentheses, subtract (x+8)
(x +7)(x +4) = 0 . . . . . . factor the equation
Solutions that make these factors zero are ...
x = -7 and x = -4
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Trying these in the original equation, we get ...
For x = -7
√(-7+8) -6 = -7
1 - 6 = -7 . . . . . . . NOT TRUE . . . extraneous solution
For x = -4
√(-4+8) -6 = -4
2 -6 = -4 . . . . . . . TRUE, actual solution
The solution is x = -4; an extraneous solution is x = -7.