Question

In simplest radical form, what are the solutions to the quadratic equation 0 =3x4x + 5?

Answer 1

Given:

The quadratic equation is

$0=-3x^2-4x+5$

To find:

The simplest radical form of the solution.

Solution:

Quadratic formula:

If a quadratic equation is $ax^2+bx+c=0$, then

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

We have,

$0=-3x^2-4x+5$

Here, a=-3, b=-4 and c=5. Putting these values in the quadratic formula, we get

$x=\dfrac{-(-4)\pm \sqrt{(-4)^2-4(-3)(5)}}{2(-3)}$

$x=\dfrac{4\pm \sqrt{16+60}}{-6}$

$x=\dfrac{4\pm \sqrt{76}}{-6}$

$x=\dfrac{4\pm 2\sqrt{19}}{-6}$

Taking 2 common, we get

$x=\dfrac{2(2\pm \sqrt{19})}{-6}$

$x=\dfrac{(2\pm \sqrt{19})}{-3}$

$x=-\dfrac{(2\pm \sqrt{19})}{3}$

Therefore, the correct option is A.