Solve (x-2)^2geq7

$$( x - 2 ) ^ { 2 } \geq 7$$

Solve for x

$$x\in (-\infty,2-\sqrt{7}]\cup [\sqrt{7}+2,\infty)$$

Solution Steps

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$\left(x-2\right)^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, -4 for b, and -3 for c in the quadratic formula.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\left(-3\right)}}{2}$$

Do the calculations.

$$x=\frac{4±2\sqrt{7}}{2}$$

Solve the equation x=\frac{4±2\sqrt{7}}{2} when ± is plus and when ± is minus.

$$x=\sqrt{7}+2 x=2-\sqrt{7}$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-\left(\sqrt{7}+2\right)\right)\left(x-\left(2-\sqrt{7}\right)\right)\geq 0$$

For the product to be ≥0, x-\left(\sqrt{7}+2\right) and x-\left(2-\sqrt{7}\right) have to be both ≤0 or both ≥0. Consider the case when x-\left(\sqrt{7}+2\right) and x-\left(2-\sqrt{7}\right) are both ≤0.

$$x-\left(\sqrt{7}+2\right)\leq 0 x-\left(2-\sqrt{7}\right)\leq 0$$

The solution satisfying both inequalities is x\leq 2-\sqrt{7}.

$$x\leq 2-\sqrt{7}$$

Consider the case when x-\left(\sqrt{7}+2\right) and x-\left(2-\sqrt{7}\right) are both ≥0.

$$x-\left(2-\sqrt{7}\right)\geq 0 x-\left(\sqrt{7}+2\right)\geq 0$$

The solution satisfying both inequalities is x\geq \sqrt{7}+2.

$$x\geq \sqrt{7}+2$$

The final solution is the union of the obtained solutions.

$$x\leq 2-\sqrt{7}\text{; }x\geq \sqrt{7}+2$$