
cosine rule
hello,
i have this
in triangle ABC, AB=(5x) 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

Re: cosine rule
You start by doing as instructed, complete the square...

Re: cosine rule
Minimum value for $y^2$ occurs at $x=\dfrac{b}{2a}=\dfrac{1}{2}$ ...

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.

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

Re: cosine rule
must be a typo ... and the smallest positive value for $y^2$ is 60.75
$y^2 = x^2  x + 61$
$y^2 = \left(x^2  x + \dfrac{1}{4}\right) + 61  \dfrac{1}{4}$
$y^2 = \left(x  \dfrac{1}{2}\right)^2 + 61  \dfrac{1}{4}$
$y^2 = \left(x  \dfrac{1}{2}\right)^2 + \dfrac{244}{4}  \dfrac{1}{4}$
$y^2 = \left(x  \dfrac{1}{2}\right)^2 + \dfrac{243}{4}$
$y_{min} = \dfrac{9\sqrt{3}}{2}$