A) -2
B) -3
C) -4
D) 2
Correct Answer: A
Solution :
Let \[\alpha \] and \[\beta \]be the zeroes of\[f\left( x \right)\]. |
Given, \[{{\alpha }^{2}}+{{\beta }^{2}}=20\] |
\[{{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta =20\] ...(i) |
For the equation, \[f\left( x \right)={{x}^{2}}-4x+k\] |
\[\alpha +\beta =4\] |
\[\alpha \beta =k\] |
Put these values in Eq. (i) |
\[\Rightarrow \,\,\,{{\left( 4 \right)}^{2}}-2k=20\] |
\[\Rightarrow \,\,16-2k=20\] |
\[\Rightarrow \,\,\,2k=-4\] |
\[\Rightarrow \,\,k=-2\] |
You need to login to perform this action.
You will be redirected in
3 sec