Solve the inequality. Write the solution set in interval notation.
-$$- x ^ { 2 } + 8 x - 13 \geq 0$$
A) $[ - 4 - \sqrt { 3 } , - 4 + \sqrt { 3 } ]$
B) $( - \infty , 4 - \sqrt { 3 } ] \cup [ 4 + \sqrt { 3 } , \infty )$
C) $( - \infty , - 4 - \sqrt { 3 } ] \cup [ - 4 + \sqrt { 3 } , \infty )$
D) $[ 4 - \sqrt { 3 } , 4 + \sqrt { 3 } ]$

Question

Quizplus · Accepted Answer

To solve the inequality $-x^2 + 8x - 13 \geq 0$, first, find the roots of the equation $-x^2 + 8x - 13 = 0$. This can be done by completing the square or using the quadratic formula. The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = -1$, $b = 8$, and $c = -13$. Plugging these values in gives $x = \frac{-8 \pm \sqrt{(8)^2 - 4(-1)(-13)}}{2(-1)} = \frac{-8 \pm \sqrt{64 - 52}}{-2} = \frac{-8 \pm \sqrt{12}}{-2} = 4 \pm \sqrt{3}$. Since the coefficient of $x^2$ is negative, the parabola opens downwards, meaning the inequality is satisfied between the roots. Therefore, the solution in interval notation is $[4 - \sqrt{3}, 4 + \sqrt{3}]$.

Solve the Inequality −x2+8x−13≥0- x ^ { 2 } + 8 x - 13 \geq 0−x2+8x−13≥0

Solve the Inequality $- x ^ { 2 } + 8 x - 13 \geq 0$