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

The answer to solvefor x,-f=sqrt(x^2-x) is x=(1+sqrt(1+4f^2))/2 {f<= 0},x=(1-sqrt(1+4f^2))/2 {f<= 0}

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solvefor x,-f=sqrt(x^2-x)

solve for

x, - f = x^{2} - x

x = \frac{1 + 1 + 4 f ^{2}}{2} {f \leq 0}, x = \frac{1 - 1 + 4 f ^{2}}{2} {f \leq 0}

Solution steps

- f = x^{2} - x

Square both sides:

f^{2} = x^{2} - x

f^{2} = x^{2} - x

Solve

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

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

Verify Solutions:

x = \frac{1 + 1 + 4 f ^{2}}{2} {f \leq 0}, x = \frac{1 - 1 + 4 f ^{2}}{2} {f \leq 0}

The solutions are

x = \frac{1 + 1 + 4 f ^{2}}{2} {f \leq 0}, x = \frac{1 - 1 + 4 f ^{2}}{2} {f \leq 0}

u = 9 u - 14

2 = 12 - x^{2}

(5 x - 4)^{\frac{1}{2}} - x = 0

2 - x = 0

5 - X + 1 = X - 2

What is the answer to solvefor x,-f=sqrt(x^2-x) ?
The answer to solvefor x,-f=sqrt(x^2-x) is x=(1+sqrt(1+4f^2))/2 {f<= 0},x=(1-sqrt(1+4f^2))/2 {f<= 0}