Question

What is the sum of the x-intercepts of the function j(x) = 6x2 + 7x - 5?

Answer 1

The given function is-

$j(x)=6x^2+7x-5$

Let's use the quadratic formula

$x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$

Where a = 6, b = 7, and c = -5.

$\begin{gathered} x=\frac{-7\pm\sqrt[]{7^2-4\cdot6\cdot(-5)}}{2\cdot6}=\frac{-7\pm\sqrt[]{49+120}}{12} \\ x=\frac{-7\pm\sqrt[]{169}}{12}=\frac{-7\pm13}{12} \\ x_1=\frac{-7+13}{12}=\frac{6}{12}=\frac{1}{2} \\ x_2=\frac{-7-13}{12}=\frac{-20}{12}=-\frac{5}{3} \end{gathered}$

Then, we add these solutions

$\frac{1}{2}-\frac{5}{3}=\frac{1\cdot3-5\cdot2}{2\cdot3}=\frac{3-10}{6}=\frac{-7}{6}$

Hence, the answer is -7/6.