If the roots of the equation  ${\mathrm{ax}}^2+\mathrm{x}+\mathrm{b}=0$ be real, then the roots of the equation ${\mathrm{x}}^2-4\sqrt{\mathrm{ab}}\mathrm{x}+1=0$ will be

${\mathrm{ax}}^2+\mathrm{x}+\mathrm{b}=0$ has real roots

$\Rightarrow {\left(1\right)}^2-4\mathrm{ab}\ge 0\Rightarrow -4\mathrm{ab}\ge 1 \mathrm{or} 4\mathrm{ab}\le 1 ...\left(\mathrm{i}\right)$

Now second equation is ${\mathrm{x}}^2-4\sqrt{\mathrm{ab}}\mathrm{x}+1=0$

Therefore $\mathrm{D}=16\mathrm{ab}-4$, from (i) $\mathrm{D}\le 0$

Hence roots are imaginary.