Solve f^-1(-5)=sqrt{2x-9}

$$f ^ { - 1 } ( - 5 ) = \sqrt { 2 x - 9 }$$

Solve for f

$$f=-\frac{5}{\sqrt{2x-9}}$$

$$x>\frac{9}{2}$$

Steps for Solving Linear Equation

Reorder the terms.

$$-5\times \frac{1}{f}=\sqrt{2x-9}$$

Variable f cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by f.

$$-5=f\sqrt{2x-9}$$

Swap sides so that all variable terms are on the left hand side.

$$f\sqrt{2x-9}=-5$$

The equation is in standard form.

$$\sqrt{2x-9}f=-5$$

Divide both sides by \sqrt{2x-9}.

$$\frac{\sqrt{2x-9}f}{\sqrt{2x-9}}=-\frac{5}{\sqrt{2x-9}}$$

Dividing by \sqrt{2x-9} undoes the multiplication by \sqrt{2x-9}.

$$f=-\frac{5}{\sqrt{2x-9}}$$

Variable f cannot be equal to 0.

$$f=-\frac{5}{\sqrt{2x-9}}\text{, }f\neq 0$$

Solve for x

$$x=\frac{9}{2}+\frac{25}{2f^{2}}$$

$$f<0$$

Steps for Solving Radical Equations

Swap sides so that all variable terms are on the left hand side.

$$\sqrt{2x-9}=f^{-1}\left(-5\right)$$

Reorder the terms.

$$\sqrt{2x-9}=-5\times \frac{1}{f}$$

Multiply both sides of the equation by f.

$$f\sqrt{2x-9}=-5$$

Divide both sides by f.

$$\frac{f\sqrt{2x-9}}{f}=-\frac{5}{f}$$

Dividing by f undoes the multiplication by f.

$$\sqrt{2x-9}=-\frac{5}{f}$$

Square both sides of the equation.

$$2x-9=\frac{25}{f^{2}}$$

Add 9 to both sides of the equation.

$$2x-9-\left(-9\right)=\frac{25}{f^{2}}-\left(-9\right)$$

Subtracting -9 from itself leaves 0.

$$2x=\frac{25}{f^{2}}-\left(-9\right)$$

Subtract -9 from \frac{25}{f^{2}}.

$$2x=9+\frac{25}{f^{2}}$$

Divide both sides by 2.

$$\frac{2x}{2}=\frac{9+\frac{25}{f^{2}}}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x=\frac{9+\frac{25}{f^{2}}}{2}$$

Divide 9+\frac{25}{f^{2}} by 2.

$$x=\frac{9}{2}+\frac{25}{2f^{2}}$$