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

The Correct Option is 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 $ .