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FP1 question is driving me mad

RIGHT, the equation, x^3+4x+3 has roots alpha beta and gamma.(a) Use the subistution x= the square root of u to obtain a cubic equation in u.This supposedly equals u^3+8u^2+16u+9=0If you can solve this n show how you will be an absolute star even my teacher couldnt

Reply 1

TeeEm

19

Original post by emilyp2206

RIGHT, the equation, x^3+4x+3 has roots alpha beta and gamma.(a) Use the subistution x= the square root of u to obtain a cubic equation in u.This supposedly equals u^3+8u^2+16u+9=0If you can solve this n show how you will be an absolute star even my teacher couldnt

maybe it is late but I cannot make sense of this .... do you have a picture of the question?

Reply 2

emilyp2206

OP

2

Its not late i need all the help give us one second

Reply 3

Notnek

21

Original post by emilyp2206

RIGHT, the equation, x^3+4x+3 has roots alpha beta and gamma.(a) Use the subistution x= the square root of u to obtain a cubic equation in u.This supposedly equals u^3+8u^2+16u+9=0If you can solve this n show how you will be an absolute star even my teacher couldnt

There's surely more to this. Can you post a picture of the question?

Edit : too late :smile:

OP

Original post by emilyp2206

RIGHT, the equation, x^3+4x+3 has roots alpha beta and gamma.(a) Use the subistution x= the square root of u to obtain a cubic equation in u.This supposedly equals u^3+8u^2+16u+9=0If you can solve this n show how you will be an absolute star even my teacher couldnt

Do you mean that you can't figure out how to get the cubic in u, or you can't go on from that?

Reply 6

emilyp2206

OP

2

Original post by notnek

There's surely more to this. Can you post a picture of the question?

Edit : too late :smile:

posted, its not too late i just left my mock corrections v. late, sorry for the bad lighting

Reply 7

emilyp2206

OP

2

Original post by C0pper

Do you mean that you can't figure out how to get the cubic in u, or you can't go on from that?

i can get as far as substituting but yeah going from there to get the cubic, ive tried rearranging and squaring but i think that might be the wrong approach

Reply 8

emilyp2206

OP

2

if anyone has a worked solution i would appreciate it SO much if you could share i picture, all of my class mates and teacher are stumped strangely

on this right now

Original post by emilyp2206

i can get as far as substituting but yeah going from there to get the cubic, ive tried rearranging and squaring but i think that might be the wrong approach

firstly I got -9 as the last term of the cubic

Secondly the cubic in u has solutions alpha², beta² and gamma²

I will now be off... hope it all makes sense

Reply 11

tinkerbella~

13

Original post by emilyp2206

i can get as far as substituting but yeah going from there to get the cubic, ive tried rearranging and squaring but i think that might be the wrong approach

Is this the 2015 paper? Swear it's the one I sat last year :lol:

I'm guessing you've got to

u \sqrt u + 4 \sqrt u + 3 = 0

just from substituting in? Try taking the 3 to the other side and squaring, it comes out pretty easily from there. Took me way longer than I'd like to admit to spot that in the actual exam.

(edited 8 years ago)

Reply 12

olegasr

6

Original post by emilyp2206

if anyone has a worked solution i would appreciate it SO much if you could share i picture, all of my class mates and teacher are stumped strangely

Dropped you a PM with the solution. :smile:

Reply 13

emilyp2206

OP

2

Original post by C0pper

Is this the 2015 paper? Swear it's the one I sat last year :lol:

I'm guessing you've got to

u \sqrt u + 4 \sqrt u + 3 = 0

just from substituting in? Try taking the 3 other to the other side and squaring, it comes out pretty easily from there. Took me way longer than I'd like to admit to spot that in the actual exam.

yeah its the last question :frown:

im still struggling

Reply 14

NinjaOtter

6

I subbed in sqrt(u) into x. Then to get a u^3 I squared the whole thing, but then I get u^3 + 8u^2 + 4u + 6u^3/2 + 24u^1/2 + 9 = 0. Also what exam board was this?

Reply 15

Astrtricks

16

shurley if you sub u into the equation it equals

u^{\frac{3}{2}}+4u^{\frac{1}{2}}+3=0

Then move the three over and square

(u^{\frac{3}{2}}+4u^{\frac{1}{2}})^2=-3^2

then expand

u^3+8u^2+16u=9

sorry for the slow reply I'm really bad at LaTeX

(edited 8 years ago)

Reply 16

atsruser

11

Original post by emilyp2206

yeah its the last question :frown:

im still struggling

u\sqrt{u}+4\sqrt{u}+3 =0 \Rightarrow \sqrt{u}(u+4)=-3 \Rightarrow ?

How can you remove the root?

Reply 17

tinkerbella~

13

Original post by NinjaOtter

I subbed in sqrt(u) into x. Then to get a u^3 I squared the whole thing, but then I get u^3 + 8u^2 + 4u + 6u^3/2 + 24u^1/2 + 9 = 0. Also what exam board was this?