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

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

Solve for x

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

Solution Steps

Square both sides of the equation.

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

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

$$16-8x+x^{2}=\left(\sqrt{x^{2}+4}\right)^{2}$$

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

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

Subtract x^{2} from both sides.

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

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

$$16-8x=4$$

Subtract 16 from both sides.

$$-8x=4-16$$

Subtract 16 from 4 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 4-x=\sqrt{x^{2}+4}.

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

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

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

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

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