Solve (x-2)^2leqx^2+1

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

Solve for x

$$x\geq \frac{3}{4}$$

Solution Steps

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

$$x^{2}-4x+4\leq x^{2}+1$$

Subtract x^{2} from both sides.

$$x^{2}-4x+4-x^{2}\leq 1$$

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

$$-4x+4\leq 1$$

Subtract 4 from both sides.

$$-4x\leq 1-4$$

Subtract 4 from 1 to get -3.

$$-4x\leq -3$$

Divide both sides by -4. Since -4 is negative, the inequality direction is changed.

$$x\geq \frac{-3}{-4}$$

Fraction \frac{-3}{-4} can be simplified to \frac{3}{4} by removing the negative sign from both the numerator and the denominator.

$$x\geq \frac{3}{4}$$