What is the answer to solvefor x,(f)x=sqrt(x^2+4) ?

The answer to solvefor x,(f)x=sqrt(x^2+4) is x= 2/(sqrt(f^2-1)) {f>1},x=-2/(sqrt(f^2-1)) {f<-1}

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solvefor x,(f)x=sqrt(x^2+4)

solve for

x, (f) x = x^{2} + 4

x = \frac{2}{f ^{2} - 1} {f > 1}, x = - \frac{2}{f ^{2} - 1} {f < - 1}

Solution steps

(f) x = x^{2} + 4

Square both sides:

x^{2} f^{2} = x^{2} + 4

x^{2} f^{2} = x^{2} + 4

Solve

x^{2} f^{2} = x^{2} + 4 : x = \frac{2}{f ^{2} - 1}, x = - \frac{2}{f ^{2} - 1}

x = \frac{2}{f ^{2} - 1}, x = - \frac{2}{f ^{2} - 1}

Verify Solutions:

x = \frac{2}{f ^{2} - 1} {f > 1}, x = - \frac{2}{f ^{2} - 1} {f < - 1}

The solutions are

x = \frac{2}{f ^{2} - 1} {f > 1}, x = - \frac{2}{f ^{2} - 1} {f < - 1}

- u + 2 u - 1 = 0

2 y = 19 y + 33

7 u - 10 = u

3 x - 1 - 2 x = 0

3 = 7 x

What is the answer to solvefor x,(f)x=sqrt(x^2+4) ?
The answer to solvefor x,(f)x=sqrt(x^2+4) is x= 2/(sqrt(f^2-1)) {f>1},x=-2/(sqrt(f^2-1)) {f<-1}