The product of the real roots of the equation | vedclass

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Question

(D) Given that $3 + i\sqrt{6}$ is a root,its conjugate $3 - i\sqrt{6}$ must also be a root since the coefficients are real.Let the other two real root

Vedclass · Accepted Answer

$-3/4$

Vedclass · Answer

(D) Given that $3 + i\sqrt{6}$ is a root,its conjugate $3 - i\sqrt{6}$ must also be a root since the coefficients are real.Let the other two real roots be $\alpha$ and $\beta$.The product of all four roots of the polynomial $ax^4 + bx^3 + cx^2 + dx + e = 0$ is given by $e/a$.Here,the product of the roots is $\alpha \cdot \beta \cdot (3 + i\sqrt{6}) \cdot (3 - i\sqrt{6}) = -45/4$.Calculating the product of the complex roots: $(3 + i\sqrt{6})(3 - i\sqrt{6}) = 3^2 + (\sqrt{6})^2 = 9 + 6 = 15$.Substituting this into the product equation: $\alpha \cdot \beta \cdot 15 = -45/4$.Therefore,$\alpha \cdot \beta = -45 / (4 	imes 15) = -45 / 60 = -3/4$.

The product of the real roots of the equation $4x^4 - 24x^3 + 57x^2 + 18x - 45 = 0$,given that one of the roots is $3 + i\sqrt{6}$,is:

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