Question

solve the quadratic inequality. state the solution as an inequality. -2x² + 15x ≥ -27
-9 ≤ x ≤ 3/2
-3/2 ≤ x ≤ 9
x ≤ -3/2 or x ≥ 9
x ≤ -9 or x ≥ 3/2

Explanation:

Step1: Rewrite the inequality

First, rewrite $-2x^{2}+15x\geq - 27$ as $-2x^{2}+15x + 27\geq0$. Let $y=-2x^{2}+15x + 27$.

Step2: Find the roots of the quadratic equation

Set $y = 0$, so $-2x^{2}+15x + 27=0$. Multiply through by - 1 to get $2x^{2}-15x - 27 = 0$. Use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $ax^{2}+bx + c = 0$. Here, $a = 2$, $b=-15$, $c=-27$. Then $x=\frac{15\pm\sqrt{(-15)^{2}-4\times2\times(-27)}}{2\times2}=\frac{15\pm\sqrt{225 + 216}}{4}=\frac{15\pm\sqrt{441}}{4}=\frac{15\pm21}{4}$. The roots are $x_1=\frac{15 + 21}{4}=9$ and $x_2=\frac{15-21}{4}=-\frac{3}{2}$.

Step3: Determine the solution of the inequality

Since $y=-2x^{2}+15x + 27$ is a parabola opening downwards (because $a=-2<0$), the inequality $-2x^{2}+15x + 27\geq0$ is satisfied when $-\frac{3}{2}\leq x\leq9$.

Answer:

$-\frac{3}{2}\leq x\leq9$