# LCM doubt in algebraic equations

I'm in a doubt on the follow equation:

Considering the equation: $x^2 + 5x - 1 = 0$, let $\alpha$ and $\beta$ be solutions; thus $\alpha*\beta = -1$ and $\alpha + \beta = -5$

Evaluate: $\dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2}$

So, working on it, I figured out that the LCM of this fraction would be $\alpha^2*\beta^2$ am I right ?

So working on it, I got:

$\dfrac{\beta^2 + \alpha^2}{\alpha^2*\beta^2}$

Is that right ? and how can I continue to evaluate it ? Thanks in advance;

Edit after @dot dot post:

So, now I've got;

$\dfrac{(\alpha + \beta)^2 - 2*\alpha*\beta}{\alpha^2*\beta^2}$

I'm done with the numerator, but what I have to do with the denominator ?

Is that possible?: $\alpha^2*\beta^2 = (\alpha*\beta)^2$ I think it's not because $2^2*3^2 \neq (2*3)^2$ What should I do now with the denominator ?

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@dotdot He doesn't know $\beta^2+\alpha^2$ yet. – Thomas Andrews Apr 5 '12 at 13:27
Are you familiar with the process of accepting answers to your questions? – Arturo Magidin Apr 5 '12 at 15:02

We can use the following reasoning $$\frac{1}{x}+\frac{1}{y}=\frac{y}{xy}+\frac{x}{xy}=\frac{x+y}{xy}$$ for any numbers $x$ and $y$ it makes sense for (i.e., as long as neither $x$ nor $y$ equals $0$). In your case, you were just using $x=\alpha^2$ and $y=\beta^2$. So your reasoning is correct.

Now, one approach would be to solve for $\alpha$ and $\beta$ explicitly using the quadratic formula, but there is a much simpler method that doesn't require that: notice that for any numbers $r$ and $s$, $$(r+s)^2=r^2+s^2+2rs,$$ and therefore $$r^2+s^2=(r+s)^2-2rs.$$ Of course, you should also know that $$r^2s^2=(rs)^2.$$ You know the values of $\alpha+\beta$ and $\alpha\beta$ - specifically, $\alpha+\beta=-1$ and $\alpha\beta=-5$. Do you see how the above formulas let you calculate $$\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\frac{\alpha^2+\beta^2}{\alpha^2\beta^2}\quad?$$

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Thank you!, now I've got it, evaluating: $\frac{(-5)^2 - 2*(-1)}{(-1)^2} = 27$ – aajjbb Apr 5 '12 at 14:00

It's right. You must apply the quadratic formula or follow Zen's hint,substitute the zeros in the expression and you are done.

The result is 27.

@aajjbb: (2*2)*(3*3) = (2*3)*(2*3)

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