What are the solutions to the equation solvefor x,f^{-1}(x)=x^2+4x ?

The solutions to the equation solvefor x,f^{-1}(x)=x^2+4x are x=0,x=-4+1/f

Find the zeros of solvefor x,f^{-1}(x)=x^2+4x

The zeros of solvefor x,f^{-1}(x)=x^2+4x are x=0,x=-4+1/f

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solve for

x, f^{- 1} (x) = x^{2} + 4 x

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

Solution steps

f^{- 1} (x) = x^{2} + 4 x

Refine

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

Switch sides

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

Move

\frac{x}{f}

to the left side

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( 4 - \frac{1}{f} ) \pm ( 4 - \frac{1}{f} ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

Simplify

(4 - \frac{1}{f})^{2} - 4 \cdot 1 \cdot 0 : - \frac{1}{f} + 4

x_{1, 2} = \frac{- ( 4 - \frac{1}{f} ) \pm ( - \frac{1}{f} + 4 )}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( 4 - \frac{1}{f} ) - \frac{1}{f} + 4}{2 \cdot 1}, x_{2} = \frac{- ( 4 - \frac{1}{f} ) - ( - \frac{1}{f} + 4 )}{2 \cdot 1}

x = \frac{- ( 4 - \frac{1}{f} ) - \frac{1}{f} + 4}{2 \cdot 1} : 0

x = \frac{- ( 4 - \frac{1}{f} ) - ( - \frac{1}{f} + 4 )}{2 \cdot 1} : - 4 + \frac{1}{f}

The solutions to the quadratic equation are:

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

- 3 x^{2} + 30 x = 0

- 0.25 x^{2} + 350 x - 120000 = 0

4 - 5 b^{2} = - 176

(x - 2)^{2} - 81 = 0

x^{2} + 25 = - 9 x + 5

What are the solutions to the equation solvefor x,f^{-1}(x)=x^2+4x ?
The solutions to the equation solvefor x,f^{-1}(x)=x^2+4x are x=0,x=-4+1/f
Find the zeros of solvefor x,f^{-1}(x)=x^2+4x
The zeros of solvefor x,f^{-1}(x)=x^2+4x are x=0,x=-4+1/f