A) 3, 5
B) 7, 9
C) 3, 6
D) 2, 5
Correct Answer: A
Solution :
Sum of zeroes of polynomial \[5{{x}^{2}}-(3+k)x+7\] is \[\frac{-[-(3+k)]}{5}\] i.e., \[\frac{3+k}{5}\] According to question, \[\frac{3+k}{5}=0\Rightarrow k=-3\] Now, \[2{{x}^{2}}-2(k+11)x+30\] becomes \[2{{x}^{2}}-16x+30.\] i.e., \[2{{x}^{2}}-16x+30=0\] or \[{{x}^{2}}-8x+15=0\] \[\Rightarrow \] \[x=3,5\] Hence, zeroes of polynomial \[2{{x}^{2}}-16x+30\] are 3,5.
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