question_answer

If \[ad\ne bc,\]then prove that the equation \[({{a}^{2}}+{{b}^{2}}){{x}^{2}}+2(ac+bd)x+({{c}^{2}}+{{d}^{2}})=0\]has no real roots.

Answer:

Given, \[ad\ne bc,\]

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

\[D={{b}^{2}}-4ac\]

\[={{[2(ac+bd)]}^{2}}-4({{a}^{2}}+{{b}^{2}})({{c}^{2}}+{{d}^{2}})\]

\[=4[{{a}^{2}}{{c}^{2}}+{{b}^{2}}{{d}^{2}}+2abcd]\]\[-4\left( {{a}^{2}}{{c}^{2}}+{{a}^{2}}{{d}^{2}}+{{b}^{2}}{{c}^{2}}+{{b}^{2}}{{d}^{2}} \right)\]

\[=4[{{a}^{2}}{{c}^{2}}+{{b}^{2}}{{d}^{2}}+2abcd-{{a}^{2}}{{c}^{2}}-{{a}^{2}}{{d}^{2}}-{{b}^{2}}{{c}^{2}}-{{b}^{2}}{{d}^{2}}]\]

\[=4[-{{a}^{2}}{{d}^{2}}-{{b}^{2}}{{c}^{2}}+2abcd]\]

\[=-4[{{a}^{2}}{{d}^{2}}+{{b}^{2}}{{c}^{2}}-2abcd]\]

\[=-4{{[ad-bc]}^{2}}\]

D is negative

Hence given equation has no real roots.                 Hence Proved.