Solve 2^2+left(8-xright)^2=x^2

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

Solve for x

$$x = \frac{17}{4} = 4\frac{1}{4} = 4.25$$

Steps for Solving Linear Equation

Calculate 2 to the power of 2 and get 4.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8-x\right)^{2}.

$$4+64-16x+x^{2}=x^{2}$$

Add 4 and 64 to get 68.

$$68-16x+x^{2}=x^{2}$$

Subtract x^{2} from both sides.

$$68-16x+x^{2}-x^{2}=0$$

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

$$68-16x=0$$

Subtract 68 from both sides. Anything subtracted from zero gives its negation.

$$-16x=-68$$

Divide both sides by -16.

$$x=\frac{-68}{-16}$$

Reduce the fraction \frac{-68}{-16} to lowest terms by extracting and canceling out -4.

$$x=\frac{17}{4}$$