Solve left(x-2right)^2>12

$${ \left(x-2 \right) }^{ 2 } > 12$$

Solve for x

$$x\in \left(-\infty,2-2\sqrt{3}\right)\cup \left(2\sqrt{3}+2,\infty\right)$$

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 -8 for c in the quadratic formula.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\left(-8\right)}}{2}$$

Do the calculations.

$$x=\frac{4±4\sqrt{3}}{2}$$

Solve the equation x=\frac{4±4\sqrt{3}}{2} when ± is plus and when ± is minus.

$$x=2\sqrt{3}+2 x=2-2\sqrt{3}$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-\left(2\sqrt{3}+2\right)\right)\left(x-\left(2-2\sqrt{3}\right)\right)>0$$

For the product to be positive, x-\left(2\sqrt{3}+2\right) and x-\left(2-2\sqrt{3}\right) have to be both negative or both positive. Consider the case when x-\left(2\sqrt{3}+2\right) and x-\left(2-2\sqrt{3}\right) are both negative.

$$x-\left(2\sqrt{3}+2\right)<0 x-\left(2-2\sqrt{3}\right)<0$$

The solution satisfying both inequalities is x<2-2\sqrt{3}.

$$x<2-2\sqrt{3}$$

Consider the case when x-\left(2\sqrt{3}+2\right) and x-\left(2-2\sqrt{3}\right) are both positive.

$$x-\left(2-2\sqrt{3}\right)>0 x-\left(2\sqrt{3}+2\right)>0$$

The solution satisfying both inequalities is x>2\sqrt{3}+2.

$$x>2\sqrt{3}+2$$

The final solution is the union of the obtained solutions.

$$x<2-2\sqrt{3}\text{; }x>2\sqrt{3}+2$$