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Questions about roots of x2 + 2xsinq + 3cos2q = 0
By Anonymous on Saturday, June 3, 2000 - 12:27 pm:

Hi, With less than 4 days to go before my A level maths exams, I really should be able to do this, and so I'm quite annoyed at myself. Please could someone help?

Find, in terms of p, the complete set of values of theta in the interval 0£q£2p for which the roots of the equation
x2 + 2xsinq + 3cos2q = 0 (1)
are real.

Now show that the roots of the equation
x2 + (5cos2q+1)x + 9cos4q = 0 (2)
are the squares of the roots of the above equation (1)

For the first part, I don't seem to be able to find the roots. There are two variables and I can't remember how to deal with these. My first decision was to change cos2q to 1-sin2q, but this still doesn't help with the x's.

n.b If anyone has the ULEAC book P2, the question is number 39 (p258). It may be easier to read on there.

Thanks very much to anyone who can help.

By James Lingard (Jchl2) on Saturday, June 3, 2000 - 02:00 pm:

OK. For the first question you need to treat the q as a constant and view the equation as a quadratic equation in x. Then you need to use the fact that the roots are real if and only if

"b2 - 4ac ³ 0"

where in this case a = 1, b = 2 sinq and c = 3 cos2q.

Therefore, the roots are real if and only if

4 sin2q - 12 cos2q ³ 0
Þ sin2q - 3 cos2q ³ 0

But sin2q = 1 - cos2q and so this is equivalent to

1 - 4 cos2q ³ 0
Þ cos2q £ 1/4
Þ -1/2 £ cosq £ 1/2

and then we need 0 £ q £ 2p, and so the required values of q are

p/3 £ q £ 2p/3 or 4p/3 £ q £ 5p/3.

If you don't see this, try drawing a graph of sinq for q in the range 0 to 2p.

By James Lingard (Jchl2) on Saturday, June 3, 2000 - 02:25 pm:

Now for the second question, we need to actually find the roots of the two equations. Again, just treat the qs as if they were constants, and solve the two equations using the quadratic formula.

James.

By Carl Evans (P2080) on Sunday, June 4, 2000 - 12:16 am:

I'm going to try it myself now, I'll post a message to let you know how I got on. I think I'll be able to solve it now.

Cheers!

By Carl Evans (P2080) on Sunday, June 4, 2000 - 01:08 am:

thats: x2+(5cos(2q)+ 1)x + 9cos4q=0

I've now done the first part for myself without difficulty, but things went a bit pear-shaped for the second part, although I understand how to do it in principle. I'll have another go in the morning! One things for sure, I'd never do it in ten minutes under exam conditions:)

By James Lingard (Jchl2) on Sunday, June 4, 2000 - 12:13 pm:

If you get stuck, do write back and I'll give you a hand.

James.

By Carl Evans (P2080) on Sunday, June 4, 2000 - 01:13 pm:

James, I've got - sinq ± (4 sin2q - 3)1/2 for the roots to the first equation, but don't seem to be able to get anything that could be the squares of the obove roots for the 2nd equation. I've given up, and would appreciate your help. I'm presuming my first roots are correct of course.

Carl

By James Lingard (Jchl2) on Sunday, June 4, 2000 - 02:25 pm:

The roots to the first equation are correct.

The way I did it is as follows:

x2 + (5 cos2q + 1)x + 9 cos4q = 0

Then seeing as you've got sin2s in your answer to the first part we'll convert the cos2s and cos4s to sin2s and sin4s. So we'll use cos2q = 1 - sin2q and cos2q = 1 - 2 sin2q, and we get

   x2 + (5(1 - 2sin2q) + 1)x + 9(1 - sin2q)2 = 0
Þ x2 + (6 - 10 sin2q)x + (9 - 18 sin2q + 9sin4q) = 0

Then using the quadratic formula,

x = ½(-(6 - 10 sin2q) ± ((6 - 10 sin2q)2 - 4(9 - 18 sin2q + 9 sin4q))½)
   = 5 sin2q - 3 ± ((9 - 30 sin2q + 25 sin4q) - (9 - 18 sin2q + 9 sin4q))½
   = 5 sin2q - 3 ± (16 sin4q - 12 sin2q)½

Then, squaring the roots to the first equation, we get

x2 = (- sinq ± (4 sin2q - 3)½)2
    = sin2q ± 2 sinq(4 sin2q - 3)½ + 4 sin2q - 3
    = 5 sin2q - 3 ± (16 sin4q - 12 sin2q)½

which are the roots to the second equation.

I hope that is clear - if not then please write back. That seems like quite a hard question for A level to me!

Good luck with the exams,
James.

By Carl Evans (P2080) on Sunday, June 4, 2000 - 10:58 pm:

James,

That makes it very clear, thanks very much. It must have taken you a while to do with the formatting, so I appreciate your help. I actually did it the same way as you, but got bogged down with the algebra and made some mistakes. However, forgive me for being a bit dim, but how do you go from 2 sinq (4 sin2q - 3)1/2 to (16 sin4q - 12 sin2q)1/2? For some reason I don't recognise this step.

Carl

By James Lingard (Jchl2) on Sunday, June 4, 2000 - 11:08 pm:

OK,

2 sinq (4 sin2q - 3)1/2
   = (4 sin2q)1/2 (4 sin2q - 3)1/2
   = ((4 sin2q)(4 sin2q - 3))1/2
   = (16 sin4q - 12 sin2q)1/2

I hope that's clear.

James.

By Carl Evans (P2080) on Monday, June 5, 2000 - 12:18 pm:

Ok, that's fine. If you're doing uni exams, good luck to you too!

Carl