# Thread: need help understanding this....

1. ## need help understanding this....

This is similar to my last post....

g over g-1 equals 8 over g+2

Ok now what I did was the following;

g(g+2) = 8(g-1)

as you can tell I cross multiplied....

$\displaystyle g^2+2g = 8g-8$

now my dilemma is about the $\displaystyle g^2$. Since I have to solve for $\displaystyle g$ the only thing I see I can do is move the the $\displaystyle 2g$ to the left and the $\displaystyle 8$ to the right so that it would look like this;

$\displaystyle g^2+8=8g-2g$

I can then combine the $\displaystyle 8g$ and the $\displaystyle 2g$ to get $\displaystyle 6g$ but what about the $\displaystyle g^2+8$....

Thanks in advance for all the help!!

2. Originally Posted by Morzilla
This is similar to my last post....

g over g-1 equals 8 over g+2

Ok now what I did was the following;

g(g+2) = 8(g-1)

as you can tell I cross multiplied....

$\displaystyle g^2+2g = 8g-8$

...
Hello,

all your considerations and calculations are OK - only the final step is missing:

$\displaystyle g^2+2g = 8g-8~\iff~g^2-6g=-8$ . Now calculate a complete square at the LHS of the equation:

$\displaystyle g^2-6g {\color{red} + 9}=-8{\color{red} + 9}~\iff~(g-3)^2=1$ . Calcualte the square-root on both sides:

$\displaystyle |\ g-3\ | = 1~\iff~g=3 \pm 1 ~\iff~g=4~\vee~g=2$

3. Originally Posted by earboth
Hello,

all your considerations and calculations are OK - only the final step is missing:

$\displaystyle g^2+2g = 8g-8~\iff~g^2-6g=-8$ . Now calculate a complete square at the LHS of the equation:

$\displaystyle g^2-6g {\color{red} + 9}=-8{\color{red} + 9}~\iff~(g-3)^2=1$ . Calcualte the square-root on both sides:

$\displaystyle |\ g-3\ | = 1~\iff~g=3 \pm 1 ~\iff~g=4~\vee~g=2$
OK two questions....

1) what is the LHS

2) where did the nine come from!?

Thanks once again!!

4. Originally Posted by Morzilla
OK two questions....

1) what is the LHS

2) where did the nine come from!?

Thanks once again!!

Hi,

1) LHS = left hand side

2) I assumed that you are familiar with the method of completing the square:
As you may know

$\displaystyle (a \pm b)^2 = a^2 \pm 2ab+b^2$

If you have

$\displaystyle g^2 - 6g$ this is the beginning of a binomial formula. You need a third summand so that you get a complete square. This 3rd summand is here $\displaystyle \left(\frac{-6}{2}\right)^2 = (-3)^2 = 9$ .
Thus

$\displaystyle g^2-6g + 9 = (g-3)^2$

5. Originally Posted by earboth
Hi,

1) LHS = left hand side

2) I assumed that you are familiar with the method of completing the square:
As you may know

$\displaystyle (a \pm b)^2 = a^2 \pm 2ab+b^2$

If you have

$\displaystyle g^2 - 6g$ this is the beginning of a binomial formula. You need a third summand so that you get a complete square. This 3rd summand is here $\displaystyle \left(\frac{-6}{2}\right)^2 = (-3)^2 = 9$ .
Thus

$\displaystyle g^2-6g + 9 = (g-3)^2$

Yes I do remember somewhat about completing the square! ahhhhh ok ok I see......Thanks!!!