Question

If b $\in {\beta }^{+}$ the roots of the equation $(2+b)x^2+(3+b)x+(4+b)=0$ is

Answer 1

$(2+b)x^2+(3+b)x+(4+b)=0$

$x=\frac{-(3+b)\pm \sqrt{(3+b{)}^2-4(4+b\left)\right(2+b)}}{2(2+b)}$

$=\frac{-(3+b)\pm \sqrt{9+b^2+6b-4(b^2+6b+8)}}{2(2+b)}$

$=\frac{-(3+b)\pm \sqrt{9+b^2+6b-4b^2-24b-32}}{2(2+b)}$

$=\frac{-(3+b)\pm \sqrt{-3b^2-18b-2}}{2(2+b)}$

it has imaginary Roots

They will be the Roots of the

$E{q}^n$.