Solve k^2+12k-4geq0

$$k ^ { 2 } + 12 k - 4 \geq 0$$

Solve for k

$$k\in (-\infty,-2\sqrt{10}-6]\cup [2\sqrt{10}-6,\infty)$$

Steps Using the Quadratic Formula

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.

$$k^{2}+12k-4=0$$

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, 12 for b, and -4 for c in the quadratic formula.

$$k=\frac{-12±\sqrt{12^{2}-4\times 1\left(-4\right)}}{2}$$

Do the calculations.

$$k=\frac{-12±4\sqrt{10}}{2}$$

Solve the equation k=\frac{-12±4\sqrt{10}}{2} when ± is plus and when ± is minus.

$$k=2\sqrt{10}-6 k=-2\sqrt{10}-6$$

Rewrite the inequality by using the obtained solutions.

$$\left(k-\left(2\sqrt{10}-6\right)\right)\left(k-\left(-2\sqrt{10}-6\right)\right)\geq 0$$

For the product to be ≥0, k-\left(2\sqrt{10}-6\right) and k-\left(-2\sqrt{10}-6\right) have to be both ≤0 or both ≥0. Consider the case when k-\left(2\sqrt{10}-6\right) and k-\left(-2\sqrt{10}-6\right) are both ≤0.

$$k-\left(2\sqrt{10}-6\right)\leq 0 k-\left(-2\sqrt{10}-6\right)\leq 0$$

The solution satisfying both inequalities is k\leq -2\sqrt{10}-6.

$$k\leq -2\sqrt{10}-6$$

Consider the case when k-\left(2\sqrt{10}-6\right) and k-\left(-2\sqrt{10}-6\right) are both ≥0.

$$k-\left(-2\sqrt{10}-6\right)\geq 0 k-\left(2\sqrt{10}-6\right)\geq 0$$

The solution satisfying both inequalities is k\geq 2\sqrt{10}-6.

$$k\geq 2\sqrt{10}-6$$

The final solution is the union of the obtained solutions.

$$k\leq -2\sqrt{10}-6\text{; }k\geq 2\sqrt{10}-6$$