A) Rational
B) Irrational
C) Real
D) Imaginary
Correct Answer: D
Solution :
\[a{{x}^{2}}+x+b=0\]has real roots Þ \[{{(1)}^{2}}-4ab\ge 0\,\,\,\Rightarrow -4ab\ge -1\]or \[4ab\le 1\] .....(i) Now second equation is \[{{x}^{2}}-4\sqrt{ab}x+1=0\] Therefore \[D=16ab-4,\]from (i) \[D\le 0\] Hence roots are imaginary.You need to login to perform this action.
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