A) -4
B) 4
C) -7
D) 7
Correct Answer: D
Solution :
| Let \[\alpha \] and \[\beta \]are the roots of given quadratic equation |
| \[{{x}^{2}}-\left( k+6 \right)x+2\left( 2k-1 \right)=0\] |
| Now, sum of roots \[=\alpha +\beta \] |
| \[=-\left\{ \frac{-\left( k+6 \right)}{1} \right\}=k+6\] |
| Product of roots \[=\alpha \beta =\frac{2\left( 2k-1 \right)}{1}=2\left( 2k-1 \right)\] |
| According to question, |
| Sum of roots (zeroes) \[=\frac{1}{2}\] |
| \[\times \] products of roots (zeroes) |
| \[\Rightarrow \,\,\,k+6=\frac{1}{2}\times 2\left( 2k-1 \right)\] |
| \[\Rightarrow \,\,\,\,\,k+6=2k-1\] |
| \[\Rightarrow \,\,\,\,\,6+1=2k-k\,\,\Rightarrow k=7\] |
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