Solve sqrt{R^2-4}+R=4

$$\sqrt { R ^ { 2 } - 4 } + R = 4$$

Solve for R

$$R = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

Solution Steps

Subtract R from both sides of the equation.

$$\sqrt{R^{2}-4}=4-R$$

Square both sides of the equation.

$$\left(\sqrt{R^{2}-4}\right)^{2}=\left(4-R\right)^{2}$$

Calculate \sqrt{R^{2}-4} to the power of 2 and get R^{2}-4.

$$R^{2}-4=\left(4-R\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-R\right)^{2}.

$$R^{2}-4=16-8R+R^{2}$$

Add 8R to both sides.

$$R^{2}-4+8R=16+R^{2}$$

Subtract R^{2} from both sides.

$$R^{2}-4+8R-R^{2}=16$$

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

$$-4+8R=16$$

Add 4 to both sides.

$$8R=16+4$$

Add 16 and 4 to get 20.

$$8R=20$$

Divide both sides by 8.

$$R=\frac{20}{8}$$

Reduce the fraction \frac{20}{8} to lowest terms by extracting and canceling out 4.

$$R=\frac{5}{2}$$

Substitute \frac{5}{2} for R in the equation \sqrt{R^{2}-4}+R=4.

$$\sqrt{\left(\frac{5}{2}\right)^{2}-4}+\frac{5}{2}=4$$

Simplify. The value R=\frac{5}{2} satisfies the equation.

$$4=4$$

Equation \sqrt{R^{2}-4}=4-R has a unique solution.

$$R=\frac{5}{2}$$