Solve a^2-68a+225leq0

$$a ^ { 2 } - 68 a + 225 \leq 0$$

Solve for a

$$a\in \begin{bmatrix}34-7\sqrt{19},7\sqrt{19}+34\end{bmatrix}$$

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.

$$a^{2}-68a+225=0$$

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, -68 for b, and 225 for c in the quadratic formula.

$$a=\frac{-\left(-68\right)±\sqrt{\left(-68\right)^{2}-4\times 1\times 225}}{2}$$

Do the calculations.

$$a=\frac{68±14\sqrt{19}}{2}$$

Solve the equation a=\frac{68±14\sqrt{19}}{2} when ± is plus and when ± is minus.

$$a=7\sqrt{19}+34 a=34-7\sqrt{19}$$

Rewrite the inequality by using the obtained solutions.

$$\left(a-\left(7\sqrt{19}+34\right)\right)\left(a-\left(34-7\sqrt{19}\right)\right)\leq 0$$

For the product to be ≤0, one of the values a-\left(7\sqrt{19}+34\right) and a-\left(34-7\sqrt{19}\right) has to be ≥0 and the other has to be ≤0. Consider the case when a-\left(7\sqrt{19}+34\right)\geq 0 and a-\left(34-7\sqrt{19}\right)\leq 0.

$$a-\left(7\sqrt{19}+34\right)\geq 0 a-\left(34-7\sqrt{19}\right)\leq 0$$

This is false for any a.

$$a\in \emptyset$$

Consider the case when a-\left(7\sqrt{19}+34\right)\leq 0 and a-\left(34-7\sqrt{19}\right)\geq 0.

$$a-\left(34-7\sqrt{19}\right)\geq 0 a-\left(7\sqrt{19}+34\right)\leq 0$$

The solution satisfying both inequalities is a\in \left[34-7\sqrt{19},7\sqrt{19}+34\right].

$$a\in \begin{bmatrix}34-7\sqrt{19},7\sqrt{19}+34\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$a\in \begin{bmatrix}34-7\sqrt{19},7\sqrt{19}+34\end{bmatrix}$$