Solve h=sqrt{x^2-(18-2x)^2/4}

$$h = \sqrt { x ^ { 2 } - \frac { ( 18 - 2 x ) ^ { 2 } } { 4 } }$$

Solve for x

$$x=\frac{h^{2}+81}{18}$$

$$h\geq 0$$

Steps for Solving Radical Equations

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(18-2x\right)^{2}.

$$h=\sqrt{x^{2}-\frac{324-72x+4x^{2}}{4}}$$

Divide each term of 324-72x+4x^{2} by 4 to get 81-18x+x^{2}.

$$h=\sqrt{x^{2}-\left(81-18x+x^{2}\right)}$$

To find the opposite of 81-18x+x^{2}, find the opposite of each term.

$$h=\sqrt{x^{2}-81+18x-x^{2}}$$

Combine x^{2} and -x^{2} to get 0.

$$h=\sqrt{-81+18x}$$

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

$$\sqrt{-81+18x}=h$$

Square both sides of the equation.

$$18x-81=h^{2}$$

Add 81 to both sides of the equation.

$$18x-81-\left(-81\right)=h^{2}-\left(-81\right)$$

Subtracting -81 from itself leaves 0.

$$18x=h^{2}-\left(-81\right)$$

Subtract -81 from h^{2}.

$$18x=h^{2}+81$$

Divide both sides by 18.

$$\frac{18x}{18}=\frac{h^{2}+81}{18}$$

Dividing by 18 undoes the multiplication by 18.

$$x=\frac{h^{2}+81}{18}$$

Divide h^{2}+81 by 18.

$$x=\frac{h^{2}}{18}+\frac{9}{2}$$

Solve for h (complex solution)

$$h=3\sqrt{2x-9}$$

Solve for x (complex solution)

$$x=\frac{h^{2}+81}{18}$$

$$arg(h)<\pi \text{ or }h=0$$

Solve for h

$$h=3\sqrt{2x-9}$$

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