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.
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