• # question_answer 4) If the roots of the equation $\left( {{a}^{2}}+{{b}^{2}} \right){{x}^{2}}-2(ac+bd)x+\left( {{c}^{2}}+{{d}^{2}} \right)=0$ are equal, prove that $\frac{a}{c}=\frac{c}{d}$.

 $\left( {{a}^{2}}+{{b}^{2}} \right){{x}^{2}}-2(ac+bd)x+\left( {{c}^{2}}+{{d}^{2}} \right)=0$ For equal roots, $D=0$ ${{[-2(ac+bd)]}^{2}}-4({{a}^{2}}+{{b}^{2}})({{c}^{2}}+{{d}^{2}})=0$ $4{{(ac+bd)}^{2}}-4({{a}^{2}}{{c}^{2}}+{{a}^{2}}{{d}^{2}}+{{b}^{2}}{{c}^{2}}+{{b}^{2}}{{d}^{2}})=0$ $4[{{a}^{2}}{{c}^{2}}+{{b}^{2}}{{d}^{2}}+2abcd-{{a}^{2}}{{c}^{2}}-{{a}^{2}}{{d}^{2}}+{{b}^{2}}{{c}^{2}}+{{b}^{2}}{{d}^{2}}]=0$ $-4[{{a}^{2}}{{d}^{2}}+{{b}^{2}}{{c}^{2}}-2abcd]=0$ $-4{{(ad-bc)}^{2}}=0$ $-4\ne 0$ so, ${{(ad-bc)}^{2}}=0$ $ad-bc=0$ $ad=bc$ $\frac{a}{b}=\frac{c}{d}$                                Hence Proved