Finding the Minimum Value of AC in a Triangle with Given Side Lengths

[tweety1234]

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.

$$x^2 -x +7 =0$$

$$x^2 -x = -7$$

$$(x- \frac{1}{2} )^2 = -7 + \frac{1}{4}$$

$$(x- \frac{1}{2} )^2 = \frac{-27}{4}$$

$$x- \frac{1}{2} = -\frac{\sqrt27}{2}$$ and than am I suppose to solve for x ? what does it mean by minimum value?

 

[tiny-tim]

$$x^2 -x +7 =0$$

uhh? where did "=0" come from??

just minimise (x - 1/2)² + 27/4 ​

 

[tweety1234]

uhh? where did "=0" come from??

just minimise (x - 1/2)² + 27/4 ​

I don't understand what you mean by 'minimise' ? $$\frac{1}{4} + \frac{27}{4} = 7$$

x=7?

 

[tiny-tim]

I don't understand what you mean by 'minimise' ?

Find the value of x for which (x - 1/2)² + 27/4 is a minimum … what is that minimum?

 

[tweety1234]

Find the value of x for which (x - 1/2)² + 27/4 is a minimum … what is that minimum?

IS the minimum 27/4 , when x = 1/2?

 

[tiny-tim]

IS the minimum 27/4 , when x = 1/2?

Yup!

That's what completing the square is all about! ​

 

[Mentallic]

$$(x- \frac{1}{2} )^2 = \frac{-27}{4}$$

$$x- \frac{1}{2} = -\frac{\sqrt27}{2}$$

This step is not correct as taking the square root of a negative number (right hand side of equation) results in complex, imaginary numbers; not the negative square root as you've done. Since x is imaginary, it means that $$x^2-x+7 \neq 0$$
And of course, you've already seen that the minimum value is 27/4 which is more than 0.