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**Flexible** Hi can someone please help me with this:

In a triangle, ABC, AB=(2-x)cm, BC=(x+1)cm and angle ABC=120;

a) show that $\displaystyle AC^2 = x^2 - x + 7$ - i did this using the cosine rule.

b) Find the value of x for which AC has a minimum value: (I need to complete the square to find x)

I get this far;

$\displaystyle (x - \frac{1}{2})^2 - (-\frac{1}{2})^2 + 7{\color{red}=0}$

$\displaystyle (x - \frac{1}{2})^2 = - \frac{27}{4}$ I took the exponent off your right hand side to make it correct.

Can you show me in stages how to resolve to find a minimum value for x?

Thanks,

F.