A) \[p{{x}^{2}}-qx+{{p}^{2}}=0\]
B) \[q{{x}^{2}}+px+{{q}^{2}}=0\]
C) \[p{{x}^{2}}+qx+{{p}^{2}}=0\]
D) \[q{{x}^{2}}-px+{{q}^{2}}=0\]
Correct Answer: B
Solution :
Given \[{{\alpha }^{3}}+{{\beta }^{3}}=-p\]and \[\alpha \beta =q\] Let\[\frac{{{\alpha }^{2}}}{\beta }\] and \[\frac{{{\beta }^{2}}}{\alpha }\]be the root of required quadratic equation. So,\[\frac{{{\alpha }^{2}}}{\beta }+\frac{{{\beta }^{2}}}{\alpha }=\frac{{{\alpha }^{2}}+{{\beta }^{3}}}{\alpha \beta }=\frac{-p}{q}\] and\[\frac{{{\alpha }^{2}}}{\beta }\times \frac{{{\beta }^{3}}}{\alpha }=\alpha \beta =q\] Hence, required quadratic equation is \[{{x}^{2}}-\left( \frac{-p}{q} \right)x+q=0\] \[\Rightarrow \]\[{{x}^{2}}+\frac{p}{q}x+q=0\Rightarrow q{{x}^{2}}+px+{{q}^{2}}=0\]
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