Question

What are the possible values of x in 8x2 + 4x = -1?

Answer 1

Answer:

The possible values of x are:

           $x=\dfrac{-1+i}{4}\ and\ x=\dfrac{-1-i}{4}$

Step-by-step explanation:

We are given a quadratic equation by:

$8x^2+4x=-1$

and we are asked to find the solution of this equation i.e the possible values of x for which this equation is true.

This equation could also be written in the form:

$8x^2+4x+1=0$

We will solve this equation by using the quadratic formula i.e. any quadratic equation of the form:

$ax^2+bx+c=0$

has a solution which is given by the formula

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

Here we have:

$a=8,\ b=4\ and\ c=1$

Hence, we have:

$x=\dfrac{-4\pm \sqrt{4^2-4\times 8\times 1}}{2\times 8}\\\\i.e.\\\\x=\dfrac{-4\pm \sqrt{16-32}}{16}\\\\i.e.\\\\x=\dfrac{-4\pm \sqrt{-16}}{16}\\\\i.e.\\\\x=\dfrac{-4\pm 4i}{16}\\\\i.e.\\\\x=\dfrac{-1\pm i}{4}$

i.e. the possible values of x are:

 $x=\dfrac{-1+i}{4}\ and\ x=\dfrac{-1-i}{4}$