If the roots of the equation ax^2 + x + b = 0 | vedclass

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(D) Given that the roots of $ax^2 + x + b = 0$ are real,the discriminant $D_1 \ge 0$.$D_1 = (1)^2 - 4ab \ge 0 \implies 4ab \le 1$.Now,consider the sec

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Imaginary

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(D) Given that the roots of $ax^2 + x + b = 0$ are real,the discriminant $D_1 \ge 0$.$D_1 = (1)^2 - 4ab \ge 0 \implies 4ab \le 1$.Now,consider the second equation $x^2 - 4\sqrt{ab}x + 1 = 0$.The discriminant $D_2$ of this equation is given by:$D_2 = (-4\sqrt{ab})^2 - 4(1)(1) = 16ab - 4$.Since $4ab \le 1$,we have $16ab \le 4$,which implies $16ab - 4 \le 0$.Therefore,$D_2 \le 0$.Since the discriminant is less than or equal to zero,the roots are imaginary.

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

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