A) \[2{{x}^{2}}-3x+5=0\]
B) \[3{{x}^{2}}-4\sqrt{3x}+4=0\]
C) \[2{{x}^{2}}-6x+3=0\]
D) \[x+\frac{1}{x}=1\]
Correct Answer: B
Solution :
(b): It is easy to mentally solve this problem. Just identify a, b, c in each of the equations. In which ever equation,\[{{b}^{2}}-4ac=0\], it will have equal roots. In (b)\[a=3\,\,\,\,\,b=-4\sqrt{3}\] and \[c=4\] \[\therefore {{b}^{2}}-4ac={{\left( -4\sqrt{3} \right)}^{2}}-4\times 3\times 4=0\]
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