A) 10
B) 20
C) 30
D) 40
E) 50
Correct Answer: D
Solution :
Let\[\alpha \]be the common root for both the equations\[{{x}^{2}}+ax+10=0\]and\[{{x}^{2}}+bx-10=0\] \[\therefore \] \[{{\alpha }^{2}}+a\alpha +10=0\] ...(i) and \[{{\alpha }^{2}}+b\alpha -10=0\] ...(ii) From Eqs. (i) and (ii), we get \[(a-b)\alpha +10+10=0\] \[\Rightarrow \] \[(a-b)\alpha =-20\] \[\Rightarrow \] \[\alpha =\frac{-20}{a-b}\] \[\because \]\[\alpha \]is also the root of\[{{x}^{2}}+bx-10=0\] \[\therefore \] \[\frac{400}{{{(a-b)}^{2}}}+b\left( \frac{-20}{a-b} \right)-10=0\] \[40-2b(a-b)={{(a-b)}^{2}}\] \[\Rightarrow \] \[{{a}^{2}}+{{b}^{2}}-2{{b}^{2}}=40\] \[\Rightarrow \] \[{{a}^{2}}-{{b}^{2}}=40\]You need to login to perform this action.
You will be redirected in
3 sec