• Sep 23rd 2010, 07:27 AM
Mbrown141
I've made it to the last question on a homework piece, but I just can't figure out how to start. The question is -

One root of the equation x² - px + q = 0 is twice the other root. Show that 2p² = 9q.

Originally, I just re-arranged 2p²=9q to p=(sqrt)9q/2, and subbed in some numbers, until I got p=4.5, q=9. But my maths tutor said that's not the answer he's looking for, and that I should be using alpha and beta, and to try again tonight. I've spent the last 45 mins scribbling bits out to no avail, I just don't know how to get the values of alpha and beta to prove 2p²=9q.

Any suggestions?
• Sep 23rd 2010, 07:31 AM
Quote:

Originally Posted by Mbrown141
I've made it to the last question on a homework piece, but I just can't figure out how to start. The question is -

One root of the equation x² - px + q = 0 is twice the other root. Show that 2p² = 9q.

Originally, I just re-arranged 2p²=9q to p=(sqrt)9q/2, and subbed in some numbers, until I got p=4.5, q=9. But my maths tutor said that's not the answer he's looking for, and that I should be using alpha and beta, and to try again tonight. I've spent the last 45 mins scribbling bits out to no avail, I just don't know how to get the values of alpha and beta to prove 2p²=9q.

Any suggestions?

let the two roots be a and b and you are told that a=2b -- 1

sum of roots, a+b = p -- 2

product of roots, ab = q -- 3

You should be able to take it from here. Just think about what you can do with the 3 simultaneous equations.
• Sep 23rd 2010, 07:33 AM
mr fantastic
Quote:

Originally Posted by Mbrown141
I've made it to the last question on a homework piece, but I just can't figure out how to start. The question is -

One root of the equation x² - px + q = 0 is twice the other root. Show that 2p² = 9q.

Originally, I just re-arranged 2p²=9q to p=(sqrt)9q/2, and subbed in some numbers, until I got p=4.5, q=9. But my maths tutor said that's not the answer he's looking for, and that I should be using alpha and beta, and to try again tonight. I've spent the last 45 mins scribbling bits out to no avail, I just don't know how to get the values of alpha and beta to prove 2p²=9q.

Any suggestions?

$(x - \alpha)(x - 2 \alpha) = x^2 - 3 \alpha x + 2 \alpha^2$.

Therefore:

$\displaystyle p = 3 \alpha \Rightarrow \frac{p}{3} = \alpha$ .... (1)

$\displaystyle q = 2 \alpha^2 \Rightarrow \frac{q}{2} = \alpha^2$ .... (2)

Square equation (1) and equate the result to equation (2).
• Sep 23rd 2010, 07:48 AM
Mbrown141
Quote:

let the two roots be a and b and you are told that a=2b -- 1

sum of roots, a+b = p -- 2

product of roots, ab = q -- 3

You should be able to take it from here. Just think about what you can do with the 3 simultaneous equations.

Already had that, my problem is not having any clue where to take it. See, I thought -

p=a+b
2p=2(a+b)
2p²=(2(a+b))²

But how do I relate that to q? Q = ab, which doesn't seem to help here. Am I taking this the complete wrong way?

Quote:

Originally Posted by mr fantastic
$(x - \alpha)(x - 2 \alpha) = x^2 - 3 \alpha x + 2 \alpha^2$.

Therefore:

$\displaystyle p = 3 \alpha \Rightarrow \frac{p}{3} = \alpha$ .... (1)

$\displaystyle q = 2 \alpha^2 \Rightarrow \frac{q}{2} = \alpha^2$ .... (2)

Square equation (1) and equate the result to equation (2).

I understand all that so far, thanks for that. Next step is to square a part of the first equation/result?

If p=3 $\alpha$, then 2p²=6 $\alpha^2$.

If q= $\frac{\alpha^2}{2}$, then 9q=4.5 $\alpha^2$, which isn't right. Not sure where I've gone wrong.
• Sep 23rd 2010, 07:53 AM
Mbrown141
Quote:

Originally Posted by mr fantastic
$(x - \alpha)(x - 2 \alpha) = x^2 - 3 \alpha x + 2 \alpha^2$.

Therefore:

$\displaystyle p = 3 \alpha \Rightarrow \frac{p}{3} = \alpha$ .... (1)

$\displaystyle q = 2 \alpha^2 \Rightarrow \frac{q}{2} = \alpha^2$ .... (2)

Square equation (1) and equate the result to equation (2).

I understand all that so far, thanks for that. Next step is to square a part of the first equation/result?

If p=3 $\alpha$, then 2p²=6 $\alpha^2$.

If q= $\frac{\alpha^2}{2}$, then 9q=4.5 $\alpha^2$, which isn't right. Not sure where I've gone wrong.
• Sep 23rd 2010, 07:56 AM
Unknown008
$\dfrac{p}{3} = \alpha$

When you square this you get:

$\dfrac{p^2}{9} = \alpha^2$

(Happy)
• Sep 23rd 2010, 08:03 AM
Mbrown141
Quote:

Originally Posted by Unknown008
$\dfrac{p}{3} = \alpha$

When you square this you get:

$\dfrac{p^2}{9} = \alpha^2$

(Happy)

So you're saying because squaring $\dfrac{p}{3} = \alpha$ = $\dfrac{p^2}{9} = \alpha^2$, then multiplying both sides by 9 gives $p^2 = 9\alpha^2$, but since $beta$ is twice $alpha$, I need to double the p²?

That would give the correct answer, just checking my understanding is right?

Thanks.

Edit: This leaves the wrong answer somehow. 2p²=9q, whereas I end up with 2p²=9q²...
• Sep 23rd 2010, 08:10 AM
Unknown008
If you look at what Mr. Fantastic gave you, he didn't include any beta. He used directly twice alpha instead of having alpha and beta.

If he used beta, then you'd have:

$(x - \alpha)(x - \beta) = x^2 - (\alpha + \beta) x + \alpha\beta$

Then;

$p = \alpha + \beta$

$q = \alpha\beta$

Then, you know that $\beta = 2\alpha$

This gives:

$p = \alpha + 2\alpha = 3 \alpha$

$q = \alpha(2\alpha) = 2\alpha^2$

Which comes back to what Mr. Fantastic gave you (Happy)
• Sep 23rd 2010, 08:13 AM
Mbrown141
Quote:

Originally Posted by Unknown008
If you look at what Mr. Fantastic gave you, he didn't include any beta. He used directly twice alpha instead of having alpha and beta.

If he used beta, then you'd have:

$(x - \alpha)(x - \beta) = x^2 - (\alpha + \beta) x + \alpha\beta$

Then;

$p = \alpha + \beta$

$q = \alpha\beta$

Then, you know that $\beta = 2\alpha$

This gives:

$p = \alpha + 2\alpha = 3 \alpha$

$q = \alpha(2\alpha) = 2\alpha^2$

Which comes back to what Mr. Fantastic gave you (Happy)

Thanks for explaining that. I understand up to that point, so I'm going to use that in the homework piece, and I know some marks are awarded for getting that far. I have no clue how to get from there to 2p²=9q, but some marks are better than none!

Thanks.
• Sep 23rd 2010, 08:20 AM
Unknown008
It's easy from there on. You can substitute the alpha in second equation, or equate alpha squared.

p + q = r
pq = 2r

Using the first equation and the second equation, I can express p and q without r.

pq = 2(p + q)

It's nearly the same thing in your problem.
• Sep 23rd 2010, 08:27 AM
Mbrown141
Quote:

Originally Posted by Unknown008
It's easy from there on. You can substitute the alpha in second equation, or equate alpha squared.

p + q = r
pq = 2r

Using the first equation and the second equation, I can express p and q without r.

pq = 2(p + q)

It's nearly the same thing in your problem.

So you're saying using the

$\alpha=\frac{p}{3}$, put that into the second equation in place of alpha, resulting in $q=\frac{2p}{3}²$. Multiplying both sides by 3, gets 3q=2p², when it needs to be 9q=2p²...

Or, do I square both top and bottom of the fraction? So I'd have q=2(( $\frac{p}{3}$)²)? That would give $q=2\frac{p^2}{9}$, then multiplying both sides by 9 would get 2p²=9q. That sounds right, anyone back me up?
• Sep 23rd 2010, 08:41 AM
Unknown008
Yep, that's it (Happy)

You square both the numerator and the denominator. The square is for both of them.