What is the answer to solvefor x,f(x-3)=sqrt(x+3)+4 ?

The answer to solvefor x,f(x-3)=sqrt(x+3)+4 is

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solvefor x,f(x-3)=sqrt(x+3)+4

solve for

x, f (x - 3) = x + 3 + 4

x = \frac{6 f ^{2} + 8 f + 1 + 24 f ^{2} + 16 f + 1}{2 f ^{2}} {f > 0}, x = \frac{6 f ^{2} + 8 f + 1 - 24 f ^{2} + 16 f + 1}{2 f ^{2}} {f \leq - \frac{2}{3}}

Solution steps

f (x - 3) = x + 3 + 4

Expand

f (x - 3) : x f - 3 f

x f - 3 f = x + 3 + 4

Remove square roots

x^{2} f^{2} - 6 x f^{2} - 8 x f + 9 f^{2} + 24 f + 16 = x + 3

Solve

x^{2} f^{2} - 6 x f^{2} - 8 x f + 9 f^{2} + 24 f + 16 = x + 3 : x = \frac{6 f ^{2} + 8 f + 1 + 24 f ^{2} + 16 f + 1}{2 f ^{2}}, x = \frac{6 f ^{2} + 8 f + 1 - 24 f ^{2} + 16 f + 1}{2 f ^{2}}

x = \frac{6 f ^{2} + 8 f + 1 + 24 f ^{2} + 16 f + 1}{2 f ^{2}}, x = \frac{6 f ^{2} + 8 f + 1 - 24 f ^{2} + 16 f + 1}{2 f ^{2}}

Verify Solutions:

x = \frac{6 f ^{2} + 8 f + 1 + 24 f ^{2} + 16 f + 1}{2 f ^{2}} {f > 0}, x = \frac{6 f ^{2} + 8 f + 1 - 24 f ^{2} + 16 f + 1}{2 f ^{2}} {f \leq - \frac{2}{3}}

The solutions are

x = \frac{6 f ^{2} + 8 f + 1 + 24 f ^{2} + 16 f + 1}{2 f ^{2}} {f > 0}, x = \frac{6 f ^{2} + 8 f + 1 - 24 f ^{2} + 16 f + 1}{2 f ^{2}} {f \leq - \frac{2}{3}}

x + x - 12 = 6

3 x + 25 = x - 5

4 x + 28 = x + 4

y = 36 - y^{2} [0.3]

x = - x - 1

What is the answer to solvefor x,f(x-3)=sqrt(x+3)+4 ?
The answer to solvefor x,f(x-3)=sqrt(x+3)+4 is