Question

What are the solutions to the quadratic equation 9x(2) + 64 = 0?

Answer 1

Answer:

Solution of given quadratic equation is

$(\frac{8i}{3}, -\frac{8i}{3} )$

Step-by-step explanation:

The given quadratic equation is

$9x^2 + 64 = 0$

The general form of the quadratic equation is given by

$ax^2 + bx + c = 0$

Comparing the general form with the given quadratic equation

$a = 9\\b = 0\\c = 64$

The solutions of the quadratic equation is given by

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

Substitute the values of a, b and c

$x=\frac{-0\pm\sqrt{0^2-4(9)(64)}}{2(9)}$

$x=\frac{\pm\sqrt{-2304}}{18}$

$x=\frac{\pm48i}{18}$

$x=\frac{\pm8i}{3}$

$x=\frac{8i}{3}$

and

$x=-\frac{8i}{3}$

Where i represents iota which means that the given quadratic equation has complex roots.

So the solution of given quadratic equation is

$(\frac{8i}{3}, -\frac{8i}{3} )$

The factored form of the given quadratic equation is

$(x+ \frac{8i}{3}) (x- \frac{8i}{3}) = 0$