question_answer

Let a, \[b\in R,\] \[a\ne 0\] be  such that the equation, \[a{{x}^{2}}-2bx+5=0\] has a repeated root \[\alpha ,\] which is also a root of the equation, \[{{x}^{2}}-2bx-10=0.\] If \[\beta \] is the other root of this equation, then \[{{\alpha }^{2}}+{{\beta }^{2}}\] is equal to: [JEE MAIN Held on 09-01-2020 Evening]

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,     \[{{\alpha }^{2}}+{{\beta }^{2}}=2(5-10a)+20\]             \[=30-20a\]             \[=25\]