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If $(2+i)$ and $(\sqrt{5}-2 i)$ are the roots of the equation $\left(x^{2}+a x+b\right)\left(x^{2}+c x+d\right)=0$ where $a, b, c$ and $d$ are real constants, then product of all the roots of the equation is
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The correct answer is:
45
If one root of a quadratic equation is of the form
$a+i b,$ then other root will be $a-i b$ So, all the roots are $2 \pm 1, \sqrt{5} \pm 2 i$
$\therefore$ Product of all the roots $=(2+i)(2-i)(\sqrt{5}+2 i)(\sqrt{5}-2 i)$
$=(4+1)(5+4)=5 \times 9=45$
$a+i b,$ then other root will be $a-i b$ So, all the roots are $2 \pm 1, \sqrt{5} \pm 2 i$
$\therefore$ Product of all the roots $=(2+i)(2-i)(\sqrt{5}+2 i)(\sqrt{5}-2 i)$
$=(4+1)(5+4)=5 \times 9=45$
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