What is the answer to solvefor x,f(x-1)=sqrt(x-4)+8 ?

The answer to solvefor x,f(x-1)=sqrt(x-4)+8 is

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solvefor x,f(x-1)=sqrt(x-4)+8

solve for

x, f (x - 1) = x - 4 + 8

x = \frac{2 f ^{2} + 16 f + 1 + - 12 f ^{2} + 32 f + 1}{2 f ^{2}} {0 < f < \frac{8 + 67}{6}}, x = \frac{2 f ^{2} + 16 f + 1 - - 12 f ^{2} + 32 f + 1}{2 f ^{2}} {\frac{8}{3} \leq f < \frac{8 + 67}{6}}

Solution steps

f (x - 1) = x - 4 + 8

Expand

f (x - 1) : x f - f

x f - f = x - 4 + 8

Remove square roots

x^{2} f^{2} - 2 x f^{2} - 16 x f + f^{2} + 16 f + 64 = x - 4

Solve

x^{2} f^{2} - 2 x f^{2} - 16 x f + f^{2} + 16 f + 64 = x - 4 : x = \frac{2 f ^{2} + 16 f + 1 + - 12 f ^{2} + 32 f + 1}{2 f ^{2}}, x = \frac{2 f ^{2} + 16 f + 1 - - 12 f ^{2} + 32 f + 1}{2 f ^{2}}

x = \frac{2 f ^{2} + 16 f + 1 + - 12 f ^{2} + 32 f + 1}{2 f ^{2}}, x = \frac{2 f ^{2} + 16 f + 1 - - 12 f ^{2} + 32 f + 1}{2 f ^{2}}

Verify Solutions:

x = \frac{2 f ^{2} + 16 f + 1 + - 12 f ^{2} + 32 f + 1}{2 f ^{2}} {0 < f < \frac{8 + 67}{6}}, x = \frac{2 f ^{2} + 16 f + 1 - - 12 f ^{2} + 32 f + 1}{2 f ^{2}} {\frac{8}{3} \leq f < \frac{8 + 67}{6}}

The solutions are

x = \frac{2 f ^{2} + 16 f + 1 + - 12 f ^{2} + 32 f + 1}{2 f ^{2}} {0 < f < \frac{8 + 67}{6}}, x = \frac{2 f ^{2} + 16 f + 1 - - 12 f ^{2} + 32 f + 1}{2 f ^{2}} {\frac{8}{3} \leq f < \frac{8 + 67}{6}}

x + 12 + x = 8

x - 6 x - 55 = 0

x, f = k \cdot x - 7

x^{2} + x - 8 = x^{2} + x + 4

4 x = 40 - 4 (68 - x^{2})

What is the answer to solvefor x,f(x-1)=sqrt(x-4)+8 ?
The answer to solvefor x,f(x-1)=sqrt(x-4)+8 is