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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 ?

  • 2
    Are you familiar with the process of [accepting answers to your questions](http://meta.math.stackexchange.com/q/3286/742)?2012-04-05

2 Answers 2

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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$2012-04-05
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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)