If (2+i) and (sqrt{5}-2i) are the roots of th | vedclass

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(C) Since the coefficients $a, b, c, d$ are real, complex roots must occur in conjugate pairs. Given roots are $z_1 = 2+i$ and $z_3 = \sqrt{5}-2i$. Th

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$45$

Vedclass · Answer

(C) Since the coefficients $a, b, c, d$ are real, complex roots must occur in conjugate pairs. Given roots are $z_1 = 2+i$ and $z_3 = \sqrt{5}-2i$. Therefore, their conjugates $z_2 = 2-i$ and $z_4 = \sqrt{5}+2i$ must also be roots of the equation. The product of all roots is given by $z_1 	imes z_2 	imes z_3 	imes z_4$. Product $= (2+i)(2-i) 	imes (\sqrt{5}-2i)(\sqrt{5}+2i)$. Using the identity $(x+iy)(x-iy) = x^2+y^2$: Product $= (2^2+1^2) 	imes ((\sqrt{5})^2+2^2) = (4+1) 	imes (5+4) = 5 	imes 9 = 45$.

If $(2+i)$ and $(\sqrt{5}-2i)$ are the roots of the equation $(x^{2}+ax+b)(x^{2}+cx+d)=0$ where $a, b, c$ and $d$ are real constants, then the product of all the roots of the equation is

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