Solve left(1-xright)^2leqleft(x+3right)^2

$${ \left(1-x \right) }^{ 2 } \leq { \left(x+3 \right) }^{ 2 }$$

Solve for x

$$x\geq -1$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-x\right)^{2}.

$$1-2x+x^{2}\leq \left(x+3\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$1-2x+x^{2}\leq x^{2}+6x+9$$

Subtract x^{2} from both sides.

$$1-2x+x^{2}-x^{2}\leq 6x+9$$

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

$$1-2x\leq 6x+9$$

Subtract 6x from both sides.

$$1-2x-6x\leq 9$$

Combine -2x and -6x to get -8x.

$$1-8x\leq 9$$

Subtract 1 from both sides.

$$-8x\leq 9-1$$

Subtract 1 from 9 to get 8.

$$-8x\leq 8$$

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

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

Divide 8 by -8 to get -1.

$$x\geq -1$$