# AS Level Further Maths Question

#1
Could anyone help with this question? Any help would be very appreciated

0
4 months ago
#2
(Original post by cloudii)
Could anyone help with this question? Any help would be very appreciated

From part b) you want to somehow square the equation (in u) to get a cubic as the expression is roughly the ~linear coefficient term. Any ideas how to get the cubic?
Last edited by mqb2766; 4 months ago
0
4 months ago
#3
It’s roots of a cubic equation, so if the roots were alpha beta and gamma the coefficients for ax^3 bx^2 … e are -b/a = alpha beta gamma , c/a = alpha beta beta gamma gamma alpha. Bc you get roots of alpha beta and gamma using root u, you can square those to compare it w the c coefficient for the value
0
#4
(Original post by mqb2766)
From part b) you want to somehow square the equation (in u) to get a cubic as the expression is roughly the ~linear coefficient term. Any ideas how to get the cubic?
No, I'm not sure on how the get the cubic. I originally tried to substitute in root u for x but realised that that wouldn't work so I got stuck.
0
4 months ago
#5
(Original post by cloudii)
No, I'm not sure on how the get the cubic. I originally tried to substitute in root u for x but realised that that wouldn't work so I got stuck.
Do the substitution. Why do you think it doesnt work / is there a slightly different way to square up the polynomial in u to get a cubic?
0
4 months ago
#6
(Original post by cloudii)
Could anyone help with this question? Any help would be very appreciated

I assume the I) is simple
So ii)Try to expand(Alphabeta+alphagamma+betagamma)^2
You should get the required term.
0
#7
(Original post by mqb2766)
Do the substitution. Why do you think it doesnt work / is there a slightly different way to square up the polynomial in u to get a cubI
I'm not sure but could you eventually say, after substituting root u and rearranging, that u^3 = (-5u-7)^2 and then expand to get that the equation is 0 = u^3 - 25u^2 - 70u - 49? Or is that not the right way to do it?
0
4 months ago
#8
(Original post by cloudii)
I'm not sure but could you eventually say, after substituting root u and rearranging, that u^3 = (-5u-7)^2 and then expand to get that the equation is 0 = u^3 - 25u^2 - 70u - 49? Or is that not the right way to do it?
Correct, so part ii) is simply the linear coeff.
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