Math Help - Completing the square.

1. Completing the square.

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

a) show that $AC^2 = x^2 - x + 7$ - 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;

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

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

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

Thanks,

F.

2. Originally Posted by Flexible

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

a) show that $AC^2 = x^2 - x + 7$ - 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;

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

$(x - \frac{1}{2})^2 = - \frac{27}{4}$ 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.
You don't have to complete the square to find the min value.

Just use $x=-\frac{b}{2a}$ in $ax^2+bx+c=0$

This will give you the x-coordinate of the vertex of your parabola.

$x^2-x+7=0$

$\frac{-(-1)}{2(1)}=\frac{1}{2}$

Now, find $f\left(\frac{1}{2}\right)$ and that will be your min value.

3. i found this video helpful on completing the square