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. ${{p}^{2}}=2q$

Answer 1

Answer: (A)

${{p}^{2}}=3q$

Answer 2

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, ${{p}^{2}}=3q$ .