Solve x^2+11x-10geq0

$${ x }^{ 2 } +11x-10 \geq 0$$

Solve for x

$$x\in (-\infty,\frac{-\sqrt{161}-11}{2}]\cup [\frac{\sqrt{161}-11}{2},\infty)$$

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.

$$x^{2}+11x-10=0$$

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, 11 for b, and -10 for c in the quadratic formula.

$$x=\frac{-11±\sqrt{11^{2}-4\times 1\left(-10\right)}}{2}$$

Do the calculations.

$$x=\frac{-11±\sqrt{161}}{2}$$

Solve the equation x=\frac{-11±\sqrt{161}}{2} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{161}-11}{2} x=\frac{-\sqrt{161}-11}{2}$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-\frac{\sqrt{161}-11}{2}\right)\left(x-\frac{-\sqrt{161}-11}{2}\right)\geq 0$$

For the product to be ≥0, x-\frac{\sqrt{161}-11}{2} and x-\frac{-\sqrt{161}-11}{2} have to be both ≤0 or both ≥0. Consider the case when x-\frac{\sqrt{161}-11}{2} and x-\frac{-\sqrt{161}-11}{2} are both ≤0.

$$x-\frac{\sqrt{161}-11}{2}\leq 0 x-\frac{-\sqrt{161}-11}{2}\leq 0$$

The solution satisfying both inequalities is x\leq \frac{-\sqrt{161}-11}{2}.

$$x\leq \frac{-\sqrt{161}-11}{2}$$

Consider the case when x-\frac{\sqrt{161}-11}{2} and x-\frac{-\sqrt{161}-11}{2} are both ≥0.

$$x-\frac{-\sqrt{161}-11}{2}\geq 0 x-\frac{\sqrt{161}-11}{2}\geq 0$$

The solution satisfying both inequalities is x\geq \frac{\sqrt{161}-11}{2}.

$$x\geq \frac{\sqrt{161}-11}{2}$$

The final solution is the union of the obtained solutions.

$$x\leq \frac{-\sqrt{161}-11}{2}\text{; }x\geq \frac{\sqrt{161}-11}{2}$$