Question

1. What are the x-intercepts ofy = 2x^2 – 3x-20?

Answer 1

The given equation is,

$y=2x^2-3x-20$

Put y=0 in the above equation and solve for x to find the x intercepts.

Putting y =0 in the above equation,

$0=2x^2-3x-20\ldots\ldots(1)_{}$

The above equation is in the form of a quadratic equation given by,

$ax^2+bx+c=0$

Comparing the equations, a=2, b=-3 and c=-20.

Now, using discriminant method solve equation (1) for x.

$\begin{gathered} x=\frac{-(-3)\pm\sqrt[]{(-3)^2-4\times2\times(-20)}}{2\times2} \\ =\frac{3\pm\sqrt[]{9^{}+160}}{4} \\ =\frac{3\pm\sqrt[]{169}}{4} \\ =\frac{3\pm13}{4} \\ x=\frac{3+13}{4}\text{ or x=}\frac{3-13}{4} \\ x=\frac{16}{4}\text{ or x=}\frac{-10}{4} \\ x=4\text{ or x=}\frac{-5}{2} \end{gathered}$

Therefore, the x intercepts of the graph of the given equation is x=4 or x=-5/2.