Question

if ab are the roots of
$x {}^{2} { + 2x + 4 = 0}$
then find
$\frac{1}{ \alpha { }^{2} } + \frac{1}{ \beta {}^{2} }$

Answer 1

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Answer 2

Answer:

$x^2+2x+4=0\:\\\:x=\frac{-2\pm \sqrt{-12}}{2}=\frac{-2\pm 2\sqrt{3}i}{2}=-1\pm \sqrt{3}i\\x=-1+\sqrt{3}i , x=-1-\sqrt{3}i$

∴$\frac{1}{\left(-1+\sqrt{3}i\right)^2}+\frac{1}{\left(-1-\sqrt{3}i\right)^2}=\frac{\left(-1-\sqrt{3}i\right)^2+\left(-1+\sqrt{3}i\right)^2}{\left(-1+\sqrt{3}i\right)^2\left(-1-\sqrt{3}i\right)^2}=\frac{\left(1+2\sqrt{3}i+3\right)+\left(1-2\sqrt{3}i-3\right)}{\left(1+2\sqrt{3}i+3\right)\left(1-2\sqrt{3}i-3\right)}=\frac{2}{\:4-12\sqrt{3}i}=\frac{2\left(4+12\sqrt{3}i\right)}{16+144\left(3\right)}=\frac{8+24\sqrt{3}i}{16+432}=\frac{8+24\sqrt{3}i}{448}=\frac{1+3\sqrt{3}i}{56}=\frac{1}{56}+\frac{3\sqrt{3}}{56}i$

Hope it's correct and hope it helps :)

Step-by-step explanation: