A) 1
B) 2
C) \[\frac{2}{3}\]
D) \[\frac{1}{3}\]
Correct Answer: D
Solution :
\[\therefore \] The roots of \[{{x}^{2}}+2(3a+5)x+2(9{{a}^{2}}+25)=0\] are equal. \[\Rightarrow \]\[{{[2(3a+5)]}^{2}}-4\times 1\times 2\left( 9{{a}^{2}}+25 \right)\] \[\Rightarrow \]\[4\left( 9{{a}^{2}}+30a+25 \right)-8\left( 9{{a}^{2}}+25 \right)=0\] \[\Rightarrow \]\[-9{{a}^{2}}+30a-25=0\] \[\Rightarrow \]\[9{{a}^{2}}-30a+25=0\] \[\Rightarrow \,\,\,\,\,\,\,\,{{\left( 3a-5 \right)}^{2}}=0\Rightarrow a=\]\[\frac{5}{3}\]
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