1. ## cosine rule

hello,
i have this

in triangle ABC, AB=(5-x) BC=(4+x) , angle ABC = 120 degrees and AC = y

i know that $\displaystyle y^2=x^2 - x + 61$

now i have to complete the square to find minimum value of y^2 and give value of x for which this occurs

but i cant figure out how to start with it, because there is no roots

2. ## Re: cosine rule

You start by doing as instructed, complete the square...

3. ## Re: cosine rule

Minimum value for $y^2$ occurs at $x=\dfrac{-b}{2a}=\dfrac{1}{2}$ ...

4. ## Re: cosine rule

I have no idea what "no roots" has to do with "completing the square". A "perfect square" is of the form $\displaystyle (x- a)^2= x^2- 2ax+ a^2$. Here, you have $\displaystyle x^2- x+ 61$. The coefficient of x is 2a= -1 so a= -1/2. Then $\displaystyle a^2= \frac{1}{4}$. To get a "perfect square" we must add (and subtract) -1/4:
[tex]x^2- x+ 61= x^2- x+ \frac{1}{4}- \frac{1}{4}+ 61= (x- \frac{1}{2})^2+ \frac{243}{244}[tex].

$\displaystyle y^2= (x- \frac{1}{2})^2+ \frac{243}{244}$.

What is the smallest value of the right side and for what value of x does that occur.

5. ## Re: cosine rule

thanks , seems i misunderstood the question somehow.. smallest value is 60.75
but what is that [tex]\frac{243}{244}[tex] ?

6. ## Re: cosine rule

must be a typo ... and the smallest positive value for $y^2$ is 60.75

$y^2 = x^2 - x + 61$

$y^2 = \left(x^2 - x + \dfrac{1}{4}\right) + 61 - \dfrac{1}{4}$

$y^2 = \left(x - \dfrac{1}{2}\right)^2 + 61 - \dfrac{1}{4}$

$y^2 = \left(x - \dfrac{1}{2}\right)^2 + \dfrac{244}{4} - \dfrac{1}{4}$

$y^2 = \left(x - \dfrac{1}{2}\right)^2 + \dfrac{243}{4}$

$y_{min} = \dfrac{9\sqrt{3}}{2}$