Question

What are the zeros of the quadratic function f(x) = 2x2 + 16x – 9? x = –4 – and x = –4 + x = –4 – and x = –4 + x = –4 – and x = –4 + x = –4 – and x = –4 +

Answer 1

We have to determine the roots of the given quadratic equation

f(x) = $2x^2+16x-9=0$

We will solve using the discriminant formula,

D = $b^2-4ac$

= $(16)^2 - 4(2)(-9)$

= $\sqrt(328)$

= $2 \sqrt2 \sqrt41$

The roots are given by:

$x= \frac{-b\pm \sqrt{D}}{2a}$

$x= \frac{-16\pm 2\sqrt{2}\sqrt{41}}{4}$

$x= -4\pm \frac{\sqrt{2}\sqrt{41}}{2}$

$x= -4\pm \frac{\sqrt{41}}{\sqrt2}$

$x= -4\pm \sqrt{\frac{41}{2}}$

Hence the roots of the given quadratic equation are $x= -4+ \sqrt{\frac{41}{2}}$ and $x= -4- \sqrt{\frac{41}{2}}$.