Find the x-intercepts.
-$$f ( x ) = 3 x ^ { 2 } + 3 x - 5$$
A) $\left( \frac { - 3 + \sqrt { 69 } } { 6 } , 0 \right)$ and $\left( \frac { - 3 - \sqrt { 69 } } { 6 } , 0 \right)$
B) $\left( \frac { 3 + \sqrt { 69 } } { 6 } , 0 \right)$ and $\left( \frac { 3 - \sqrt { 69 } } { 6 } , 0 \right)$
C) $\left( \frac { 3 + \sqrt { 51 } } { 6 } , 0 \right)$ and $\left( \frac { 3 - \sqrt { 51 } } { 6 } , 0 \right)$
D) $\left( \frac { - 3 + \sqrt { 51 } } { 6 } , 0 \right)$ and $\left( \frac { - 3 - \sqrt { 51 } } { 6 } , 0 \right)$

Question

Quizplus · Accepted Answer

To find the x-intercepts of the quadratic equation $f(x) = 3x^2 + 3x - 5$, we use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 3$, $b = 3$, and $c = -5$. Plugging these values in, we get $x = \frac{-3 \pm \sqrt{(3)^2 - 4(3)(-5)}}{2(3)} = \frac{-3 \pm \sqrt{9 + 60}}{6} = \frac{-3 \pm \sqrt{69}}{6}$, which matches choice A.

Find the X-Intercepts f(x)=3x2+3x−5f ( x ) = 3 x ^ { 2 } + 3 x - 5f(x)=3x2+3x−5

Find the X-Intercepts $f ( x ) = 3 x ^ { 2 } + 3 x - 5$