<s>[INST] $\left(\frac1\alpha-\frac1\beta\right)^2$ for $p(x)=x^2+x-2$ If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $p(x)=x^2+x-2$,
then $\left(\frac1\alpha-\frac1\beta\right)^2 is:$
A) $\frac94$
B) $\frac{-9}4$
C) $\frac25$
D) $\frac{-2}5$
This is a homework question from our school's home assignment for class 10. I tried this:
We can write $p(x)=x^2+x-2$ as $$p(x)=\underbrace{(1)x^2}_{ax^2}
+\underbrace{(1)x}_{bx}+\underbrace{(2)}_c$$ Therefore, $$a=1\quad
b=1\quad c=2$$ We know that $$\alpha\beta=\frac ca=2$$ and
$$\alpha+\beta=\frac{-b}{a}=-1$$ Now $$\left(\frac1\alpha-\frac1
\beta\right)^2=\left(\frac{\alpha-\beta}{\alpha\beta}\right)^2=
\frac{(\alpha-\beta)^2}{(\alpha\beta)^2}=\frac{\alpha^2+\beta^2-2
\alpha\beta}4=\frac{\alpha^2+\beta^2-4}4$$
I got stuck here. What to do now?
 [/INST] $(x+\frac{1}{2})^2-\frac{1}{4} -2 =0 \iff (x+\frac{1}{2})^2=\frac{9}{4} \iff x=1$ or $x=-2$
Since we know that $\alpha$ and $\beta$ are the roots of the polynomial, it must hold that $\alpha =1$ and $\beta=-2$ or $\alpha=-2$ and $\beta=1$
Lets first consider the case in which $\alpha =1$ and $\beta=-2.  \Rightarrow (\frac{1}{\alpha}-\frac{1}{\beta})^2=(1+\frac{1}{2})^2=\frac{9}{4}$
In the case in which $\alpha=-2$ and $\beta=1 \Rightarrow (\frac{1}{\alpha}-\frac{1}{\beta})^2=(-\frac{1}{2}-1)^2=\frac{9}{4}$, so we arrive at the same result.
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