question_answer

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 \[\frac{a}{c}=\frac{c}{d}\].

Answer:

\[\left( {{a}^{2}}+{{b}^{2}} \right){{x}^{2}}-2(ac+bd)x+\left( {{c}^{2}}+{{d}^{2}} \right)=0\] 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