A) \[{{p}^{2}}=3q\]
B) \[{{p}^{2}}>3q\]
C) \[{{p}^{2}}<3q\]
D) \[{{p}^{2}}=2q\]
E) \[p=3q\]
Correct Answer: A
Solution :
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,\[{{p}^{2}}=3q\].
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