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

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

Solve for x

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{x^{2}+4} to the power of 2 and get x^{2}+4.

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

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

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

Add 8x to both sides.

$$x^{2}+4+8x=16+x^{2}$$

Subtract x^{2} from both sides.

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

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

$$4+8x=16$$

Subtract 4 from both sides.

$$8x=16-4$$

Subtract 4 from 16 to get 12.

$$8x=12$$

Divide both sides by 8.

$$x=\frac{12}{8}$$

Reduce the fraction \frac{12}{8} to lowest terms by extracting and canceling out 4.

$$x=\frac{3}{2}$$

Substitute \frac{3}{2} for x in the equation \sqrt{x^{2}+4}=4-x.

$$\sqrt{\left(\frac{3}{2}\right)^{2}+4}=4-\frac{3}{2}$$

Simplify. The value x=\frac{3}{2} satisfies the equation.

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

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

$$x=\frac{3}{2}$$