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basic algebra

by vaibhav101 » Wed May 16, 2018 8:20 am

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if a and b are the roots of the equation $$x^2+x+2=0$$ ,then $$\frac{\left(a^{10}+b^{10}\right)}{\left(a^{-10}+b^{-10}\right)}$$ is
A 4096
B 2048
C 1024
D 512
E 256
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by Vincen » Fri May 18, 2018 4:35 am
Hello vaibhav101.

I think there is a mistake in the equation. It should be x^2+x-2=0 instead of x^2+x+2=0.

Using the first equation we get:

First, let's find the roots of $$x^2+x-2=0.$$ $$\Rightarrow\ \ x=\frac{-1\pm\sqrt{1+8}}{2}=\frac{-1\pm\sqrt{9}}{2}=\frac{-1\pm 3}{2}$$ This implies that: $$a=\frac{-1+3}{2}=1\ \ \ \ and\ \ \ \ \ \ b=\frac{-1-3}{2}=-2.$$ Hence

$$a^{10}=1\ \ \ and\ \ \ \ \ a^{-10}=1.$$ $$b^{10}=1024\ \ \ and\ \ \ \ \ b^{-10}=\frac{1}{1024}.$$ Now, $$\frac{a^{10}+b^{10}}{a^{-10}+b^{-10}}=\frac{1+1024}{1+\frac{1}{1024}}=\frac{1025}{\frac{1025}{1024}}=1024.$$ Therefore, the correct answer is the option C.

I hope it helps.
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