A) \[{{x}^{2}}-2abx-{{({{a}^{2}}-{{b}^{2}})}^{2}}=0\]
B) \[{{x}^{2}}-4abx-{{({{a}^{2}}-{{b}^{2}})}^{2}}=0\]
C) \[{{x}^{2}}-4abx+{{({{a}^{2}}-{{b}^{2}})}^{2}}=0\]
D) None of these
Correct Answer: B
Solution :
Sum of roots \[\alpha +\beta =-(a+b)\]and \[\alpha \beta =\frac{{{a}^{2}}+{{b}^{2}}}{2}\] Þ \[{{(\alpha +\beta )}^{2}}={{(a+b)}^{2}}\]and \[{{(\alpha -\beta )}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}-2\alpha \beta \] = \[2ab-({{a}^{2}}+{{b}^{2}})=-{{(a-b)}^{2}}\] Now the required equation whose roots are \[{{(\alpha +\beta )}^{2}}\]and \[{{(\alpha -\beta )}^{2}}\] \[{{x}^{2}}-\{{{(\alpha +\beta )}^{2}}+{{(\alpha -\beta )}^{2}}\}\,x+{{(\alpha +\beta )}^{2}}{{(\alpha -\beta )}^{2}}=0\] \[\Rightarrow {{x}^{2}}-\{{{(a+b)}^{2}}-{{(a-b)}^{2}}\}\,x-{{(a+b)}^{2}}{{(a-b)}^{2}}=0\] \[\Rightarrow {{x}^{2}}-4abx-{{({{a}^{2}}-{{b}^{2}})}^{2}}=0\]
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