Use the quadratic formula or graphing methods to
-$$4 x ^ { 2 } + 20 x + 8 \geq 0$$
A) $x \geq \frac { - 5 + \sqrt { 17 } } { 2 } \approx - 0.438$
B) $\frac { - 5 - \sqrt { 17 } } { 2 } \approx - 4.562 \leq x \leq \frac { - 5 + \sqrt { 17 } } { 2 } \approx - 0.438$
C) $x \leq \frac { - 5 - \sqrt { 17 } } { 2 } \approx - 4.562$
D) $x \leq \frac { - 5 - \sqrt { 17 } } { 2 } \approx - 4.562$ or $x \geq \frac { - 5 + \sqrt { 17 } } { 2 } \approx - 0.438$

Question

Quizplus · Accepted Answer

The quadratic equation $4x^2 + 20x + 8 \geq 0$ can be solved using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 4$, $b = 20$, and $c = 8$. Solving gives the roots as $\frac{-5 \pm \sqrt{17}}{2}$. Since it's a "greater than or equal to" inequality and the coefficient of $x^2$ is positive (indicating a parabola opening upwards), the solution includes values outside the interval between the roots, leading to $x \leq \frac{-5 - \sqrt{17}}{2}$ or $x \geq \frac{-5 + \sqrt{17}}{2}$.

Use the Quadratic Formula or Graphing Methods to - 4x2+20x+8≥04 x ^ { 2 } + 20 x + 8 \geq 04x2+20x+8≥0

Use the Quadratic Formula or Graphing Methods to
- $4 x ^ { 2 } + 20 x + 8 \geq 0$