Quadratic Function Bounds for β: Solving for β in Terms of α, a, and b

[kalupahana]

Homework Statement

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

Deduce that, If 12b²< 49ac< 49b²/2

then β lies between 3α/4 and 4α/3

Homework Equations

α+β = -b/a

αβ = c/a

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.