Find the value of ##r## and ##s## in the given quadratic equation

[chwala]

Homework Statement

see attached

Relevant Equations

quadratics

Let the roots of the given quadratic equation be $x=α$ and $x=β$ then our quadratic equation will be of the form;
$$x^2-(α+β)x+αβ$$
It follows that $(α+β)=(r+is)$ and $αβ=4$.
We are informed that $α^2+β^2=6i$ then $$6i=(r+is)(r+is)-8$$ ... $$8+6i=(r^2-s^2)+2rsi$$
solving the simultaneous equation yields,
$rs=3$ giving us $s$=$\dfrac {3}{r}$
$r^4-8r^2-9=0$. Let $m$=$r^2$
$m^2-8m-9=0$
$m=9$, $m=-1$
but we know that $m=r^2$ therefore $r^2=9, ⇒r=±3$ also $r^2=-1$ (unsuitable)
then we shall end up with $r=3, s=1$ and $r=-3, s=-1$

Any other way of doing it...highly appreciated ...

 

[PeroK]

Homework Statement:: see attached
Relevant Equations:: quadratics

View attachment 296997Let the roots of the given quadratic equation be $x=α$ and $x=β$ then our quadratic equation will be of the form;
$$x^2-(α+β)x+αβ$$
It follows that $(α+β)=(r+is)$ and $αβ=4$.
We are informed that $α^2+β^2=6i$ then $$6i=(r+is)(r+is)-8$$

At this point, with $z = r + is$, you have
$$z^2 = 8 + 6i$$$$z = \pm\sqrt{8 + 6i}$$