The difference between the roots of the equat | vedclass

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(A) For a quadratic equation $ax^2 + bx + c = 0$, the roots are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.Here, $a = 1, b = 2, c = 4$.The root

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$2\sqrt{3}$

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(A) For a quadratic equation $ax^2 + bx + c = 0$, the roots are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.Here, $a = 1, b = 2, c = 4$.The roots are $x = \frac{-2 \pm \sqrt{2^2 - 4(1)(4)}}{2(1)} = \frac{-2 \pm \sqrt{4 - 16}}{2} = \frac{-2 \pm \sqrt{-12}}{2}$.Since $\sqrt{-12} = \sqrt{12} \cdot i = 2\sqrt{3}i$, the roots are $x = \frac{-2 \pm 2\sqrt{3}i}{2} = -1 \pm \sqrt{3}i$.Let the roots be $\alpha = -1 + \sqrt{3}i$ and $\beta = -1 - \sqrt{3}i$.The difference between the roots is $|\alpha - \beta| = |(-1 + \sqrt{3}i) - (-1 - \sqrt{3}i)| = |2\sqrt{3}i| = 2\sqrt{3}$.

The difference between the roots of the equation $x^2 + 2x + 4 = 0$ is ... (where $i = \sqrt{-1}$)

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