Finding the Value of n for One Root to be the Reciprocal of the Other

[Porty]

This is probably very easy, but anyway..

I just don't really understand the question, therefore don't know what answer to give..

Homework Statement

The question reads...

For what values of $$n$$ will one root of the equation
$$(n - 2)x^2 + (n + 2)x + 2n + 1 = 0$$ be the reciprocal of the other?

I know the answer is $$n = -3$$

Homework Equations

if $$\alpha$$ and $$\beta$$ are the roots of $$ax^2 + bx + c = 0,$$
then $$\alpha + \beta = -\frac{b}{a}$$ and $$\alpha\beta = \frac{c}{a}$$

The Attempt at a Solution

I do
$$\alpha + \beta = - \frac{(n+2)}{(n-2)}$$

then solve for $$n$$ which gives me $$n=2$$

and

$$\alpha\beta = \frac{(2n+1)}{(n-2)}$$

then solve for $$n$$ which gives me $$n=-3$$

I need to know how the $$n=-3$$ relates to the answer..??

 

[songoku]

$$\alpha + \beta = - \frac{(n+2)}{(n-2)}$$

then solve for $$n$$ which gives me $$n=2$$

Can you give an explanation how you got n = 2?
The value n= 2 will make the denominator zero

$$\alpha\beta = \frac{(2n+1)}{(n-2)}$$

then solve for $$n$$ which gives me $$n=-3$$

Reciprocal roots : the roots are $$\alpha$$ and $$\frac{1}{\alpha}$$

So, $$(\alpha)(\frac{1}{\alpha}) = \frac{(2n+1)}{(n-2)}$$

and yes, n = -3.

So n = -3 implies that the roots are reciprocal ($$\alpha$$ and $$\frac{1}{\alpha}$$)

 

[Дьявол]

What the task says is:

$$\alpha=1/\beta$$

And you got

$$\alpha + \frac{1}{\alpha} = - \frac{(n+2)}{(n-2)}$$
and

$$\alpha*\frac{1}{\alpha}=\frac{(2n+1)}{(n-2)}$$Just solve it, and you got the solution.

Regards.

 

[Porty]

Thanks for the replies guys..
Just quickly leting you guys know why I am studying this..
I hope to get into uni next year, since i didn't do year 11 and 12 i decided to pick up a book on 2 unit math to prepare myself for what may be ahead... So some of my questions may seem really easy...

Ok i think i get it..

for the product of the roots $$\alpha * \frac{1}{\alpha}$$ simplifies to $$1$$ it becomes..

$$1 = \frac{(2n+1)}{(n-2)}$$

And you get $$n = -3$$

But what do you do with the sum of the roots..
$$\alpha + \frac{1}{\alpha} = -\frac{(n+2)}{(n-2)}$$

 

[Gib Z]

Well, basically, you do have two bits of information, the bit with the product of the roots, and the bit with the sum of the roots. Now, the question doesn't ask anything specific about each root themselves, just their product, which is why you only have to use the product of the roots equation.

Using that alone is enough to solve your problem, without going into what the actual roots are themselves. If you wanted to actually solve the roots themselves you could use the sum of the roots equation. As you know n=-3 now, a+ 1/a = -1/5.

Taking the constant over and multiplying by a, you can get a quadratic equation that you can solve. Note however this is unnecessary, as since we know n=-3, but can put that back into the original equation you started with, and it is the same equation as the one we just made.

Just out of interest, do you live in NSW, Australia? I just gather than as not many other places use the "2 unit" system.

 

[Porty]

Good guess... Yep NSW Australia.. haha

Yeah i knew that i did't have to figure out the roots themselves to solve it so i wasn't going to attempt it.. Your right, using the products of the roots did end up being enough to solve the product of the roots equation. I did'n even think of substituting $$n$$ back into the sum of the roots.. Do i need to..??

Just quickly.. What if you were given the value of one root but not the other, and were asked to find $$n$$..? Something like...

$$\alpha + 4 = n + 2$$ and $$\alpha * 4 = n - 4$$

Would i need to do it the long way and figure out the value of the other root or can it be done simpler..? I hope there is enough information there...

 

[Gib Z]

Do you didn't need to sub n back into the sum of roots, I was just saying that if you did you could find the roots themselves, which was unnecessary for this question.

For your second question, is really depends on what you know. If you know the precise equation, ie x^2 + 5x+ 6, and given that one root was -3, you could use the product of roots formula to get that the other root is -2.

In this particular equation, you don't know the values of the coefficients specifically, so you could only find the other root in terms of n and not as a number.

 

[Porty]

Do you didn't need to sub n back into the sum of roots, I was just saying that if you did you could find the roots themselves, which was unnecessary for this question.

Oh ok.Thanks..

Um ok here's what i got the second equation from...

The question is:

For what value of $$k$$ will the equation $$x^2 - (k+2)x + (k-4)=0$$ have:
a) one root equal to zero.
b) one root equal to 4?
c) one root which is the reciprocal of the other?

a) was easy
c) was easy (thanks to this forum)

But b) i can't solve.. The answer is $$k = 1\frac{1}{3}$$

I get to:

$$\alpha + 4 = k + 2$$ or $$\alpha * 4 = k - 4$$

And then don't know where to go..? I am probably just looking at it the wrong way..??

Sorry if it seems i need my hand held through this... lol

 

[Gib Z]

Ahh ok. Well, Good work, those are the two bits of info you needed to extract. Now you have to solve them simultaneously- ie, find a solution that satisfies both the equations.

To do that, we have to get all the alphas to be replaced with k's so we can solve for k's. So choose one of the equations, make alpha the subject, and replace alpha with that in the other equation, then solve.

Good luck!

 

[Porty]

ha ha.. Thank you! That works sweet... I wish i could think of doing stuff like that on my own...

anyways here it is..

I took the sum:
$$\alpha + 4 = k + 2$$
$$\alpha = k - 2$$

and put it in the product:
$$\alpha * 4 = k - 4$$

and it gave me exactly what i needed...

$$4(k-2) = k - 4$$

I luv it when it all clicks...