Solve the inequality. Write the solution set in interval notation.
-$- x ^ { 2 } + 6 x - 4 \geq 0$
A) $( \infty , - 3 - \sqrt { 5 } ] \cup [ - 3 + \sqrt { 5 } , \infty )$
B) $( \infty , 3 - \sqrt { 5 } ] \cup [ 3 + \sqrt { 5 } , \infty )$
C) $[ - 3 - \sqrt { 5 } , - 3 + \sqrt { 5 } ]$
D) $[ 3 - \sqrt { 5 } , 3 + \sqrt { 5 } ]$

Question

Quizplus · Accepted Answer

To solve the inequality $-x^2 + 6x - 4 \geq 0$, we first find the roots of the equation $-x^2 + 6x - 4 = 0$. Factoring or using the quadratic formula, we get the roots as $x = 3 \pm \sqrt{5}$. Since the coefficient of $x^2$ is negative, the parabola opens downwards, meaning the values between the roots satisfy the inequality. Thus, the solution in interval notation is $[3 - \sqrt{5}, 3 + \sqrt{5}]$.

Solve the Inequality −x2+6x−4≥0- x ^ { 2 } + 6 x - 4 \geq 0−x2+6x−4≥0

Solve the Inequality $- x ^ { 2 } + 6 x - 4 \geq 0$