Solve r+sqrt{r^2-36^2}=96

$$r + \sqrt { r ^ { 2 } - 36 ^ { 2 } } = 96$$

Solve for r

$$r = \frac{219}{4} = 54\frac{3}{4} = 54.75$$

Solution Steps

Subtract r from both sides of the equation.

$$\sqrt{r^{2}-36^{2}}=96-r$$

Calculate 36 to the power of 2 and get 1296.

$$\sqrt{r^{2}-1296}=96-r$$

Square both sides of the equation.

$$\left(\sqrt{r^{2}-1296}\right)^{2}=\left(96-r\right)^{2}$$

Calculate \sqrt{r^{2}-1296} to the power of 2 and get r^{2}-1296.

$$r^{2}-1296=\left(96-r\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(96-r\right)^{2}.

$$r^{2}-1296=9216-192r+r^{2}$$

Add 192r to both sides.

$$r^{2}-1296+192r=9216+r^{2}$$

Subtract r^{2} from both sides.

$$r^{2}-1296+192r-r^{2}=9216$$

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

$$-1296+192r=9216$$

Add 1296 to both sides.

$$192r=9216+1296$$

Add 9216 and 1296 to get 10512.

$$192r=10512$$

Divide both sides by 192.

$$r=\frac{10512}{192}$$

Reduce the fraction \frac{10512}{192} to lowest terms by extracting and canceling out 48.

$$r=\frac{219}{4}$$

Substitute \frac{219}{4} for r in the equation r+\sqrt{r^{2}-36^{2}}=96.

$$\frac{219}{4}+\sqrt{\left(\frac{219}{4}\right)^{2}-36^{2}}=96$$

Simplify. The value r=\frac{219}{4} satisfies the equation.

$$96=96$$

Equation \sqrt{r^{2}-1296}=96-r has a unique solution.

$$r=\frac{219}{4}$$