Solve x^2-4x-5leq0

$$x ^ { 2 } - 4 x - 5 \leq 0$$

Solve for x

$$x\in \begin{bmatrix}-1,5\end{bmatrix}$$

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.

$$x^{2}-4x-5=0$$

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 -5 for c in the quadratic formula.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\left(-5\right)}}{2}$$

Do the calculations.

$$x=\frac{4±6}{2}$$

Solve the equation x=\frac{4±6}{2} when ± is plus and when ± is minus.

$$x=5 x=-1$$

Rewrite the inequality by using the obtained solutions.

$$\left(x-5\right)\left(x+1\right)\leq 0$$

For the product to be ≤0, one of the values x-5 and x+1 has to be ≥0 and the other has to be ≤0. Consider the case when x-5\geq 0 and x+1\leq 0.

$$x-5\geq 0 x+1\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-5\leq 0 and x+1\geq 0.

$$x+1\geq 0 x-5\leq 0$$

The solution satisfying both inequalities is x\in \left[-1,5\right].

$$x\in \begin{bmatrix}-1,5\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-1,5\end{bmatrix}$$