Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question
1  
@dotdot He doesn't know $\beta^2+\alpha^2$ yet. –  Thomas Andrews Apr 5 '12 at 13:27
2  
Are you familiar with the process of accepting answers to your questions? –  Arturo Magidin Apr 5 '12 at 15:02
add comment

2 Answers 2

up vote 2 down vote accepted

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 $$\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\frac{\alpha^2+\beta^2}{\alpha^2\beta^2}\quad?$$

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

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)

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.