Question

I just need to make sure 1-9 are correct so if you could confirm please.9 b)

Answer 1

Given:

$2x^2+3x+9=0$

To find:

The roots.

Explanation:

Here,

$\begin{gathered} a=2 \\ b=3 \\ c=9 \end{gathered}$

Using the quadratic formula,

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

On substitution we get,

$\begin{gathered} x=\frac{-3\pm\sqrt{(3)^2-4(2)(9)}}{2(2)} \\ =\frac{-3\pm\sqrt{9-72}}{4} \\ =\frac{-3\pm\sqrt{-63}}{4} \\ =\frac{-3\pm3i\sqrt{7}}{4} \end{gathered}$

Therefore, the solutions are,

$x=\frac{-3+3\imaginaryI\sqrt{7}}{4},x=\frac{-3-3\imaginaryI\sqrt{7}}{4}$

Final answer:

The solutions are,

$x=\frac{-3+3\imaginaryI\sqrt{7}}{4},x=\frac{-3-3\imaginaryI\sqrt{7}}{4}$