Find the x-intercepts.
-$$f ( x ) = 2 x ^ { 2 } - 3 x - 4$$
A) $\left( \frac { 3 + \sqrt { 41 } } { 4 } , 0 \right)$ and $\left( \frac { 3 - \sqrt { 41 } } { 4 } , 0 \right)$
B) No x-intercepts
C) $\left( \frac { 3 + \sqrt { 29 } } { 4 } , 0 \right)$ and $\left( \frac { 3 - \sqrt { 29 } } { 4 } , 0 \right)$
D) $( 0,0 )$ and $( 4,0 )$

Question

Quizplus · Accepted Answer

To find the x-intercepts, we need to set $f(x)$ to 0 and solve for $x$.
\begin{align*}
2x^2-3x-4 &= 0 \
x &= \frac{-b\pm\sqrt{b^2-4ac}}{2a} \
x &= \frac{3\pm\sqrt{(-3)^2-4(2)(-4)}}{2(2)} \
x &= \frac{3\pm\sqrt{41}}{4}
\end{align*}
Therefore, the x-intercepts are at $\left(\frac{3+\sqrt{41}}{4}, 0ight)$ and $\left(\frac{3-\sqrt{41}}{4}, 0ight)$. The answer is choice A.

Find the X-Intercepts f(x)=2x2−3x−4f ( x ) = 2 x ^ { 2 } - 3 x - 4f(x)=2x2−3x−4

Find the X-Intercepts $f ( x ) = 2 x ^ { 2 } - 3 x - 4$