Question

Let $z_1$ and $z_2$ be the roots of the equation $z^2+pz+q=0$ where p, q are real. The points represented by $z_1,z_2$ and the origin form an equilateral triangle, if

• A. $p^2=3q$
• B. $p^2>3q$
• C. $p^2<3q$
• D. $p^2=2q$
• E. $p=3q$

Answer 1

Answer: (A)

We have, $z^2+pz+q=0$ and let $p^2=3q$
$\Rightarrow$ $z=\frac{-p\pm \sqrt{p^2-4q}}{2}=\frac{-p\pm \sqrt{3q-4q}}{2}$
$=\frac{-p\pm i\sqrt{q}}{2}$ Let $z_1=\frac{-p+i\sqrt{q}}{2}$ and $z_2=\frac{-p-i\sqrt{q}}{2}$
Further, let $z_1$ and $z_2$ be the affixes of points A and B respectively. Then, $OA=|z_1|=\sqrt{{\left(-\frac{p}{2}\right)}^2+{\left(\frac{\sqrt{q}}{2}\right)}^2}=\sqrt{\frac{p^2}{4}+\frac{q}{4}}$
$=\sqrt{\frac{3q}{4}+\frac{q}{4}}=\sqrt{q}$ $OB=|z_2|=\sqrt{{\left(-\frac{p}{2}\right)}^2+{\left(-\frac{\sqrt{q}}{2}\right)}^2}=\sqrt{\frac{p^2}{4}+\frac{q}{4}}$
$=\sqrt{\frac{3q}{4}+\frac{q}{4}}=\sqrt{q}$ and $AB=|z_1-z_2|=|i\sqrt{q}|=\sqrt{0+{(\sqrt{q})}^2}$
$=\sqrt{q}$
$\therefore$ $OA=OB=AB$
$\Rightarrow$ $\Delta AOB$ is an equilateral triangle. Thus, $p^2=3q$ .