FP1 Polynomials - finding equations with roots

This question gives it in terms of a fraction, and normally i'd use U = (root) and rearrange to get the root on its own in terms of U, but when its a fraction they're two roots involved and im quite sure what to do.
Part iii

This question gives it in terms of a fraction, and normally i'd use U = (root) and rearrange to get the root on its own in terms of U, but when its a fraction they're two roots involved and im quite sure what to do.
Part iii

EDIT - i have just figured, could you re arrange to get Ua = b
Ua + a = b + a , b + a = -5
Ua + a = -5
a(u+1) = -5
a = -5/(u+1) ?

Forget this U stuff, the easy way to remember is a + B = -x coefficient/x^2 coefficient
and aB=constant/x^2 coefficient

Ok after much effort I think the answer to (iii) is 10x^2 - 5x + 10. I hate typing out working but could do it if you really want. Basically if the roots are a/B and B/a the sum of them will = -x cof/x^2 cof and a/B + B/a = (a^2 + B^2)/aB from the last question we know these values are 5 and 10 so an example equation would be -x cof = 5, hence x cof = -5, x^2 cof = 10 and then the product of B/a and a/B = the constant/ cof X^2 (which we know can be 10 when cof x = -5)
a/B * B/a = 1 therefore the constant/ cof X^2 = 10 therefore the constant is also 10.
I hope this is right.
P.s. Alex Jones is a moron - but please don't let this sentence affect your decision to rep :P

answers say x^2 - 1/2x +1 :/

He's a wise man

Yes and mine is just that answer multiplied by 10 so they would still have the same solutions and I would still get the marks. As long as the negative x cof/x^2 cof = -1/2 and the constant and x^2 cof are equal the solution holds

This is the way I would do part 3:
We know, ab=10, and a+b=-5, and a^2+b^2=5 from the previous parts.
Having roots as a/b and b/a means that (a/b)*(b/a)=C (with an equation of the form Ax^2+Bx+C). Multiplying this together you get ab/ab, which =1, hence C=1
Similarly (a/b)+(b/a)=B. Multiplying to get the common denominator ab, we get (a^2+b^2)/ab. Since a^2+b^2=5, and ab=10, B=-5/10 (-1/2). Therefore we get the equation x^2-(1/2)x+1.

Same answer as above poster, just slightly different method

i know that ab = c, but how does ab/ba also = C? surely it would always equal 1?

Sorry, I think I explained it badly... makes sense in my head

Say g and h represent the roots of any equation. This means gh=c. With part three of the question, you are told that the roots are a/b and b/a. So for this particular equation, let g=a/b and h=b/a. gh is now (a/b)*(b/a)=ab/ba. So ab/ba=c for these particular roots, which in this case means that c does =1, hence the x^2-(1/2)x+1