If alpha, beta are the irrational roots of th | vedclass

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(C) Given the equation $3p^2x^3 + px^2 + qx + 3 = 0$. Substituting $p = 1$ and $q = -7$,we get: $3x^3 + x^2 - 7x + 3 = 0$. By testing $x = 1$,we find

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$\frac{2\sqrt{13}}{3}$

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(C) Given the equation $3p^2x^3 + px^2 + qx + 3 = 0$. Substituting $p = 1$ and $q = -7$,we get: $3x^3 + x^2 - 7x + 3 = 0$. By testing $x = 1$,we find $3(1)^3 + (1)^2 - 7(1) + 3 = 3 + 1 - 7 + 3 = 0$. Thus,$(x - 1)$ is a factor. Dividing the polynomial by $(x - 1)$,we get: $(x - 1)(3x^2 + 4x - 3) = 0$. The roots are $x = 1$ and the roots of $3x^2 + 4x - 3 = 0$. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get: $x = \frac{-4 \pm \sqrt{16 - 4(3)(-3)}}{2(3)} = \frac{-4 \pm \sqrt{16 + 36}}{6} = \frac{-4 \pm \sqrt{52}}{6} = \frac{-4 \pm 2\sqrt{13}}{6} = \frac{-2 \pm \sqrt{13}}{3}$. Let $\alpha = \frac{-2 + \sqrt{13}}{3}$ and $\beta = \frac{-2 - \sqrt{13}}{3}$. Then $|\alpha - \beta| = |\frac{-2 + \sqrt{13} - (-2 - \sqrt{13})}{3}| = |\frac{2\sqrt{13}}{3}| = \frac{2\sqrt{13}}{3}$.

If $\alpha, \beta$ are the irrational roots of the equation $3p^2x^3 + px^2 + qx + 3 = 0$ when $p = 1$ and $q = -7$,then $|\alpha - \beta| = $

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