Question

if a and b are the roots of the equation 2x^2 - px + 7 =0 , a/b is a root of the equation

(a) 7x^2 + 28x - p^2x + 14 =0

(b) 14x^2 + (28-p^2)x + 14 =0

Answer 1

from what the roots are we can infer that\[(2x-2a)(x-b)=0\]from multiplying it out and combining like terms we know that\[-2b-2a=-p\]and the final coefficient is\[2ab=7\]

Answer 2

\[2a+2b=p\]\[ab=\frac72\]

Answer 3

we can eliminate at least one of the variables at this point

Answer 4

\[ab=7\implies b=\frac7{2a}\]put this into the first equation\[2a+\frac7a=p\]\[2a^2+7a-pa=0\]\[a[2a+7-p]=0\]hm... still not drawing any conclusions

Answer 5

\(a\) can't be zero from the first line in the above post, so that leaves that\[2a+7-p=0\]\[a={7-p\over2}\]we could have done the same argument with b, so that seems to mean that\[a=b=\frac{7-p}2\]

Answer 6

somebody feel free to criticize what I'm doing or jump in or whatever

Answer 7

you got:\[2a+2b=p\]and:\[2ab=7\]therefore:\[2(a+b)=p\implies4(a+b)^2=p^2\implies4(a^2+b^2+2ab)=p^2\]therefore:\[4(a^2+b^2+7)=p^2\implies a^2+b^2=\frac{p^2}{4}-7\]

Answer 8

now, lets say that the other quadratic has a roots of a/b and say r, and that it is of the form:\[x^2-ux+v=0\]then this implies:\[\frac{ar}{b}=u\]and\[\frac{a}{b}+r=v\]

Answer 9

this leads to:\[r=\frac{ub}{a}\]

Answer 10

therefore:\[\frac{a}{b}+\frac{ub}{a}=v\]multiplying both sides by ab gives:\[a^2+ub^2=vab\]

Answer 11

the first equation:\[14x^2 + (28-p^2)x + 14 =0\]can be divided by 14 to give:\[x^2-(\frac{p^2}{14}-2)x+1=0\]which gives:\[v=\frac{p^2}{14}\]and:\[u=1\]

Answer 12

sorry I had u and v the wrong way around up there

Answer 13

oh .. simply replace 'p' by 'a, b' in those options ... and try to factor out (x-a/b) from those options

Answer 14

I should have initially said the equation is:\[x^2-vx+u=0\]

Answer 15

this then gives:\[a^2+ub^2=vab\]leading to:\[a^2+b^2=(\frac{p^2}{14}-2)ab=(\frac{p^2}{14}-2)\times\frac{7}{2}=\frac{p^2}{4}-7\]

Answer 16

which meets the original equation we had for \(a^2+b^2\)

Answer 17

I made /some/ algebra mistakes up there but hopefully it all adds up correctly

Answer 18

@Yahoo! did you follow that or do you want to explain anything more?

Answer 19

*want me to...

Answer 20

you have
right??