A) 25
B) 24
C) 26
D) 28
Correct Answer: A
Solution :
The given equations are \[a{{x}^{2}}-2bx+5=0\] and \[{{x}^{2}}-2bx-10=0\] and \[4{{b}^{2}}=20a\Rightarrow \,\,{{b}^{2}}=5a\] Now, \[{{\alpha }^{2}}+{{\beta }^{2}}={{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta \] \[=4{{b}^{2}}+20\] As \['\alpha '\] is a root of \[{{x}^{2}}-2bx-10=0\] \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\alpha }^{2}}-2b\alpha =10\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{5}{a}-2b\cdot \frac{b}{a}=10\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5-2{{b}^{2}}=10a\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5-10a=10a\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a=\frac{1}{4}\] Now, \[{{\alpha }^{2}}+{{\beta }^{2}}=2(5-10a)+20\] \[=30-20a\] \[=25\]You need to login to perform this action.
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