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.
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?
If p=3 , then 2p²=6 .
If q= , then 9q=4.5 , which isn't right. Not sure where I've gone wrong.
That would give the correct answer, just checking my understanding is right?
Edit: This leaves the wrong answer somehow. 2p²=9q, whereas I end up with 2p²=9q²...
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:
Then, you know that
Which comes back to what Mr. Fantastic gave you
It's easy from there on. You can substitute the alpha in second equation, or equate alpha squared.
Like if you had
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.
, put that into the second equation in place of alpha, resulting in . 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(( )²)? That would give , then multiplying both sides by 9 would get 2p²=9q. That sounds right, anyone back me up?