Quadratika!-II

We have the equation

$\color{#D61F06}{x^2+5x+8=0}$

we know that for a quadratic equation of the form $ax^2+bx+c=0$

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$ ,using this our equation has

$x=\dfrac{-5\pm \sqrt{25-4(8)}}{2(1)} = \dfrac{-5}{2} \pm \dfrac{\sqrt{7}}{2} i$

so $a= \dfrac{-5}{2}$ and $b=\dfrac{\sqrt{7}}{2} \implies a+b^2=\frac{-5}{2}+\frac{7}{4} = \dfrac{-3}{4} = \boxed{-0.75}$

The solutions to $x^2 + 5x + 8 = 0$ are a pair of complex conjugates $a+bi$ and $a-bi$ . By Vieta's formula , we have:

$\begin{cases} a+bi + a - bi = - 5 & \implies a = - \frac 52 & \\ (a+bi)(a-bi) = 8 & \implies a^2+b^2 = 8 & \implies b^2 = 8 - \frac {25}4 = \frac 74 \end{cases}$

$\implies a+b^2 = - \frac 52 + \frac 74 = - \frac 34 = \boxed{-0.75}$

The solutions are complex conjugates: $0=(x-(a+bi))(x-(a-bi))=x^2-2ax+(a^2+b^2).$ Comparing with the equation that was given, we find $-2a=5\Longrightarrow a=-2\tfrac{1}{2};$ $a^2+b^2=8\Longrightarrow b^2=8-a^2=8-6\tfrac{1}{4}=1\tfrac{3}{4}.$ From there is is clear that $a+b^2=-2\tfrac{1}{2}+1\tfrac{3}{4}=-\tfrac{3}{4}.$