Solve x^2-16x+16>0^2

$${ x }^{ 2 } -16x+16 > { 0 }^{ 2 }$$

Solve for x

$$x\in \left(-\infty,8-4\sqrt{3}\right)\cup \left(4\sqrt{3}+8,\infty\right)$$

Steps Using the Quadratic Formula

Calculate 0 to the power of 2 and get 0.

$$x^{2}-16x+16>0$$

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.

$$x^{2}-16x+16=0$$

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, -16 for b, and 16 for c in the quadratic formula.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 1\times 16}}{2}$$

Do the calculations.

$$x=\frac{16±8\sqrt{3}}{2}$$

Solve the equation x=\frac{16±8\sqrt{3}}{2} when ± is plus and when ± is minus.

$$x=4\sqrt{3}+8 x=8-4\sqrt{3}$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-\left(4\sqrt{3}+8\right)\right)\left(x-\left(8-4\sqrt{3}\right)\right)>0$$

For the product to be positive, x-\left(4\sqrt{3}+8\right) and x-\left(8-4\sqrt{3}\right) have to be both negative or both positive. Consider the case when x-\left(4\sqrt{3}+8\right) and x-\left(8-4\sqrt{3}\right) are both negative.

$$x-\left(4\sqrt{3}+8\right)<0 x-\left(8-4\sqrt{3}\right)<0$$

The solution satisfying both inequalities is x<8-4\sqrt{3}.

$$x<8-4\sqrt{3}$$

Consider the case when x-\left(4\sqrt{3}+8\right) and x-\left(8-4\sqrt{3}\right) are both positive.

$$x-\left(8-4\sqrt{3}\right)>0 x-\left(4\sqrt{3}+8\right)>0$$

The solution satisfying both inequalities is x>4\sqrt{3}+8.

$$x>4\sqrt{3}+8$$

The final solution is the union of the obtained solutions.

$$x<8-4\sqrt{3}\text{; }x>4\sqrt{3}+8$$