Question

The quadratic formula gives which roots for the equation 2x^2 + x - 6 = 0?

Answer 1

Answer:

The roots are $x = (\frac{3}{2}, -2)$, which is given by option C.

Step-by-step explanation:

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$.

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$, given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}$

$\Delta = b^{2} - 4ac$

In this question, we have that:

$2x^2 + x - 6 = 0$

Which is a quadratic equation with $a = 2, b = 1, c = -6$. So

$\Delta = 1^{2} - 4*2(-6) = 1 + 48 = 49$

$x_{1} = \frac{-1 + \sqrt{49}}{2*2} = \frac{6}{4} = \frac{3}{2}$

$x_{2} = \frac{-1 - \sqrt{49}}{2*2} = \frac{-8}{4} = -2$

So the roots are $x = (\frac{3}{2}, -2)$, which is given by option C.