Solve x^2-8x+16+(x+0)^2leq2x^2

$$x ^ { 2 } - 8 x + 16 + ( x + 0 ) ^ { 2 } \leq 2 x ^ { 2 }$$

Solve for x

$$x\geq 2$$

Solution Steps

Anything plus zero gives itself.

$$x^{2}-8x+16+x^{2}\leq 2x^{2}$$

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

$$2x^{2}-8x+16\leq 2x^{2}$$

Subtract 2x^{2} from both sides.

$$2x^{2}-8x+16-2x^{2}\leq 0$$

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

$$-8x+16\leq 0$$

Subtract 16 from both sides. Anything subtracted from zero gives its negation.

$$-8x\leq -16$$

Divide both sides by -8. Since -8 is negative, the inequality direction is changed.

$$x\geq \frac{-16}{-8}$$

Divide -16 by -8 to get 2.

$$x\geq 2$$