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) |
| \[\Rightarrow \,\,\,{{\left( 4 \right)}^{2}}-2k=20\] |
| \[\Rightarrow \,\,16-2k=20\] |
| \[\Rightarrow \,\,\,2k=-4\] |
| \[\Rightarrow \,\,k=-2\] |
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