Question

The roots of an equation are x = -1 ± i.
The equation is x^2 + _____ + 2 = 0.

Answer 1

"The roots of an equation are x = -1 ± i. "

so the eqn is (x-(-1+i))*(x-(-1-i)) = 0

expend the above n simply...what will u get?

Answer 2

Let p and q be the roots of \(\Large x^2 + bx + c = 0\)


Rule: \(\Large p+q = -b\) so \(\Large b = -(p+q)\)

Answer 3

well if you think about the general quadratic formula

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

you can see from the equation a = 1 and c = 2 so

\[x = \frac{-b \pm \sqrt{b^2 - 4 \times 1 \times 2}}{2 \times 1}\]

then comparing the parts

\[-1 = \frac{-b}{2}\]

solve for b...

and then check with b by using the rest of the quadratic formula

\[b^2 - 8 = \frac{\sqrt{4i^2}}{2}\]