You are Here: Home >< Maths

1. Ive just done this stupid sheet for further maths and the answers at the bottom are different to what i got and its driving me nuts, someone please give me the correct solutions so i understand..

The Roots of the equation 4x^2+8x+1=0 are a (alpha) and B (beta), find the values of

i) a^3*B + a*B^3

ii) B^2/a + a^2/B

Thanks if anyone can help
2. I get the roots of the equation to be root(3)/2 -1 and -root(3)/2 -1

And I get i to be 7/8 and ii to be -26.

Out of interest, what did you get?
3. Quadratic of the form ax²+bx+c has solutions (-b + √(b² - 4ac))/2a and (-b - √(b² - 4ac))/2a

Call the first solution alpha, and the second one beta.

Then alpha is -1 + √(0.75) and beta is -1 - √(0.75)

And I get

i) = 7/8
ii) = -26
4. (Original post by AntiMagicMan)
I get the roots of the equation to be root(3)/2 -1 and -root(3)/2 -1

And I get i to be 7/8 and ii to be -26.

Out of interest, what did you get?
Ahh, just beat me.
5. Ah, nevermind, reassurance that I am right always helps .

Even doing a maths degree is no guarantee that you are going to be right

If the poster got the same as us, then it is probally safe to assume the question sheet answers are wrong, (that was a frequent occurance with my maths teachers, they did it to make sure we weren't cheating... sneaky of them ).
6. (Original post by mr_tomus)
Ive just done this stupid sheet for further maths and the answers at the bottom are different to what i got and its driving me nuts, someone please give me the correct solutions so i understand..

The Roots of the equation 4x^2+8x+1=0 are a (alpha) and B (beta), find the values of

i) a^3*B + a*B^3

ii) B^2/a + a^2/B

Thanks if anyone can help
Going the long way round,

4x^2+8x+1 = 0
roots are [-8 +/- sqrt(64-16)]/8 = +/- sqrt48 - 1
Call a = sqrt48 - 1
Call B = -sqrt48 - 1

a^3*B + a*B^3 = ab(aa + bb) = (-49)(48+1-2.sqrt48 + 48+1+2.sqrt48)
(-49)(98) = -4802?

that seems way off... maybe its right i just dont have a feeling that is right
7. That should be sqrt(48)/8 i.e. root(3)/2

See the two answers above anyway.
8. (Original post by AntiMagicMan)
If the poster got the same as us, then it is probally safe to assume the question sheet answers are wrong, (that was a frequent occurance with my maths teachers, they did it to make sure we weren't cheating... sneaky of them ).
Our textbooks had mistakes in the answers section, but my teachers didn't know where they were, so sometimes they were defending the book when it was wrong.
9. Hehe, yep sounds about normal. Our chemistry teacher shouldn't really have been called a teacher, merely a narrator for the textbook .
10. Because 4x^2 + 8x+1 factorises as 4(x-a)(x-b) = 4(x^2 -(a+b)x+ab) then we can see that a+b = -8/4 = -2 and ab = 1/4.

a^3 b + b^3 a = ab(a^2+b^2) = ab((a+b)^2-2ab) = 1/4 x (4 - 1/2) = 7/8

b^2/a+a^2/b = (b^3 + a^3)/ab = ((a+b)^3 - 3ab(a+b))/ab

= (-8 + 3/2)/(1/4) = -26

can both be expressed in terms of a+b and ab as they're symmetric.

So you don't actually need to calculate a and b. This methods generalises to higher degree polys where you can't even find the roots.
11. Yes that is a good method, especially useful for quartics and higher.
12. Cheers people, im not dumb after all! i got the same as u and i know some answers on the sheet are wrong already so i guess these must be too. thanks

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: August 5, 2004
Today on TSR

### Are you living on a tight budget at uni?

From budgets to cutbacks...

### University open days

1. University of Cambridge
Wed, 26 Sep '18
2. Norwich University of the Arts
Fri, 28 Sep '18
3. Edge Hill University
Faculty of Health and Social Care Undergraduate
Sat, 29 Sep '18
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams