- #1
tweety1234
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Homework Statement
In △ABC, AB=(2-x) cm, BC=(x+1) cm and ∠ABC=120°:
(a) Show that AC^2=x^2-x+7.
(b) Find the value of x for which AC has a minimum value
The Attempt at a Solution
I am having trouble with part 'b'
am suppose to complete the square , but I don't seem to get the right answer.
[tex] x^2 -x +7 =0 [/tex]
[tex] x^2 -x = -7 [/tex]
[tex] (x- \frac{1}{2} )^2 = -7 + \frac{1}{4} [/tex]
[tex](x- \frac{1}{2} )^2 = \frac{-27}{4} [/tex]
[tex] x- \frac{1}{2} = -\frac{\sqrt27}{2}[/tex] and than am I suppose to solve for x ? what does it mean by minimum value?