10th Class Mathematics Solved Paper - Mathematics 2017 Outside Delhi Set-III

  • 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 \[\frac{a}{c}=\frac{c}{d}\].

    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


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