Completing the Square Question, thanks.

[Steve Macwell]

Hello this is the first time I've used this forum. I'm a little stuck with how to go about using the 'Completing the Square method' to answer the following question:

6S^2*c1*r1*c2*r2 + S(3c1*r1 + 2c2 *r2)+1 = 0

Thanks for anyone who can offer any insight into how to go about solving this problem via completing the square.

Thanks I really appreciate it.

 

[HallsofIvy]

Hello this is the first time I've used this forum. I'm a little stuck with how to go about using the 'Completing the Square method' to answer the following question:

6S^2*c1*r1*c2*r2 + S(3c1*r1 + 2c2 *r2)+1 = 0

Thanks for anyone who can offer any insight into how to go about solving this problem via completing the square.

Thanks I really appreciate it.

Since the only square I see is "\(\displaystyle S^2\)", I assume you are trying to solve for S. Is that correct?

The basic idea in "completing the square" is that \(\displaystyle (x+ a)^2= x^2+ 2ax+ a^2\) so that to have a "perfect square", the constant term (\(\displaystyle a^2\)) must be the square of half the coefficient of x (2a). Also note that this has nothing multiplied by \(\displaystyle x^2\) so the first thing I would recommend doing is dividing both sides by 6c1*r1*c2*r2 (or \(\displaystyle 6c_1r_1c_2r_2\)) leaving \(\displaystyle S^2+ \frac{3c_1r_1+ 2c_2r_2}{6c_1r_1c_2r_2}S+ \frac{1}{6c_1r_1c_2r_2}= 0\).

I would also recommend moving that constant term to the right (subtracting it from both sides) so you can focus on completing the square on the left:
\(\displaystyle S^2+ \frac{3c_1r_1+ 2c_2r_2}{6c_1r_1c_2r_2}S= -\frac{1}{6c_1r_1c_2r_2}\)
Now, what is 1/2 the coefficient of S? What is the square of that? Add that square to both sides.

You will be able to write the left sides as \(\displaystyle (S+ \frac{3c_1r_1+ 2c_2r_2}{12c_1r_1c_2r_2})^2\)
and then take the square root of both sides.

 

[Deleted member 4993]

Hello this is the first time I've used this forum. I'm a little stuck with how to go about using the 'Completing the Square method' to answer the following question:

6S^2*c1*r1*c2*r2 + S(3c1*r1 + 2c2 *r2)+1 = 0

Thanks for anyone who can offer any insight into how to go about solving this problem via completing the square.

Thanks I really appreciate it.

I double beg you to start this way:

A = 6S^2*c1*r1*c2*r2

B = (3c1*r1 + 2c2 *r2)

C = 1

Then your equation becomes:

AS^2 + B*S + C = 0

AS^2 + B*S = -C

S^2 + B/A*S = -C/A

S^2 + 2*{B/(2A)}*S + {B/(2A)}^2 = {B/(2A)}^2 - C/A

[S + B/(2A)]^2 = {B/(2A)}^2 - C/A

Square has been completed.......

 

[Steve Macwell]

Well I'll most certainly return to this glorious forum with more math dilemmas. Thanks a lot guys. Much appreciated!!