Quadratic function problem

[kalupahana]

1. The problem statement, all variables and given/known data

..........................49ac-12b2
(4α-3β)(3α-4β) =------------------------
....................................a2

Deduce that, If 12b2< 49ac< 49b2/2

then β lies between 3α/4 and 4α/3

2. Relevant equations

α+β = -b/a

αβ = c/a

3. The attempt at a solution

[kalupahana]

I don't know how to do this.

I have even a idea to about these type of question, please help me

[Mentallic]

Well, start by expanding the left hand side of that equality. Use the related equations you've given and remember the fact that (a+b)^2=a^2+b^2+2ab

[kalupahana]

Well, start by expanding the left hand side of that equality. Use the related equations you've given and remember the fact that (a+b)^2=a^2+b^2+2ab

12α2-25αβ + 12β2

Using this i got that this in terms of a, b & c.

Next part of the question is this. How should i do it

[Mentallic]

Right so looking at your a2 and b2 part, if (a+b)^2=a^2+b^2+2ab then a^2+b^2=(a+b)^2-2ab

[kalupahana]

12α2 + 12β2 = (12α+12β)2 - 313αβ

[Mentallic]

No not quite. If you expanded that you would get (12a)^2+(12b)^2=144a^2+144b^2

12a^2+12b^2=12(a^2+b^2)=12((a+b)^2-2ab)

Now go on from this.