Solve (k-1)^2-16geq0

$$( k - 1 ) ^ { 2 } - 16 \geq 0$$

Solve for k

$$k\in (-\infty,-3]\cup [5,\infty)$$

Steps Using the Quadratic Formula

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

$$k^{2}-2k+1-16\geq 0$$

Subtract 16 from 1 to get -15.

$$k^{2}-2k-15\geq 0$$

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.

$$k^{2}-2k-15=0$$

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, -2 for b, and -15 for c in the quadratic formula.

$$k=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 1\left(-15\right)}}{2}$$

Do the calculations.

$$k=\frac{2±8}{2}$$

Solve the equation k=\frac{2±8}{2} when ± is plus and when ± is minus.

$$k=5 k=-3$$

Rewrite the inequality by using the obtained solutions.

$$\left(k-5\right)\left(k+3\right)\geq 0$$

For the product to be ≥0, k-5 and k+3 have to be both ≤0 or both ≥0. Consider the case when k-5 and k+3 are both ≤0.

$$k-5\leq 0 k+3\leq 0$$

The solution satisfying both inequalities is k\leq -3.

$$k\leq -3$$

Consider the case when k-5 and k+3 are both ≥0.

$$k+3\geq 0 k-5\geq 0$$

The solution satisfying both inequalities is k\geq 5.

$$k\geq 5$$

The final solution is the union of the obtained solutions.

$$k\leq -3\text{; }k\geq 5$$