A) Real roots
B) Purely imaginary roots
C) Imaginary roots
D) None of these
Correct Answer: A
Solution :
Let \[{{D}_{1}}\] and \[{{D}_{2}}\] be discriminants of \[{{x}^{2}}+{{b}_{1}}x+{{c}_{1}}=0\] and \[{{x}^{2}}+{{b}_{2}}x+{{c}_{2}}=0\] respectively. Then \[{{D}_{1}}+{{D}_{2}}=b_{1}^{2}-4{{c}_{1}}+b_{2}^{2}-4{{c}_{2}}=(b_{1}^{2}+b_{2}^{2})-4({{c}_{1}}+{{c}_{2}})\] \[b_{1}^{2}+b_{2}^{2}-2{{b}_{1}}{{b}_{2}}\,\,,\,\,\,\,\,\,\,\,\,\,\,(\,\,\because \,\,{{b}_{1}}{{b}_{2}}=2({{c}_{1}}+{{c}_{2}})\,)\] = \[{{({{b}_{1}}-{{b}_{2}})}^{2}}\ge 0\] Þ \[{{D}_{1}}\ge 0\]or \[{{D}_{2}}\ge 0\]or \[{{D}_{1}}\]and \[{{D}_{2}}\]both are positive.
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