Solve n^2+24n-24^2leq0

$$n ^ { 2 } + 24 n - 24 ^ { 2 } \leq 0$$

Solve for n

$$n\in \begin{bmatrix}-12\sqrt{5}-12,12\sqrt{5}-12\end{bmatrix}$$

Steps Using the Quadratic Formula

Calculate 24 to the power of 2 and get 576.

$$n^{2}+24n-576\leq 0$$

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.

$$n^{2}+24n-576=0$$

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, 24 for b, and -576 for c in the quadratic formula.

$$n=\frac{-24±\sqrt{24^{2}-4\times 1\left(-576\right)}}{2}$$

Do the calculations.

$$n=\frac{-24±24\sqrt{5}}{2}$$

Solve the equation n=\frac{-24±24\sqrt{5}}{2} when ± is plus and when ± is minus.

$$n=12\sqrt{5}-12 n=-12\sqrt{5}-12$$

Rewrite the inequality by using the obtained solutions.

$$\left(n-\left(12\sqrt{5}-12\right)\right)\left(n-\left(-12\sqrt{5}-12\right)\right)\leq 0$$

For the product to be ≤0, one of the values n-\left(12\sqrt{5}-12\right) and n-\left(-12\sqrt{5}-12\right) has to be ≥0 and the other has to be ≤0. Consider the case when n-\left(12\sqrt{5}-12\right)\geq 0 and n-\left(-12\sqrt{5}-12\right)\leq 0.

$$n-\left(12\sqrt{5}-12\right)\geq 0 n-\left(-12\sqrt{5}-12\right)\leq 0$$

This is false for any n.

$$n\in \emptyset$$

Consider the case when n-\left(12\sqrt{5}-12\right)\leq 0 and n-\left(-12\sqrt{5}-12\right)\geq 0.

$$n-\left(-12\sqrt{5}-12\right)\geq 0 n-\left(12\sqrt{5}-12\right)\leq 0$$

The solution satisfying both inequalities is n\in \left[-12\sqrt{5}-12,12\sqrt{5}-12\right].

$$n\in \begin{bmatrix}-12\sqrt{5}-12,12\sqrt{5}-12\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$n\in \begin{bmatrix}-12\sqrt{5}-12,12\sqrt{5}-12\end{bmatrix}$$