need help understanding this....

**Morzilla** · December 11th 2007, 06:14 PM

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....

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

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

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

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

Thanks in advance for all the help!!

**earboth** · December 11th 2007, 09:12 PM

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....

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

...

Hello,

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

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

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

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

**Morzilla** · December 12th 2007, 03:25 AM

Originally Posted by earboth

Hello,

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

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

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

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

OK two questions....

1) what is the LHS

2) where did the nine come from!?

Thanks once again!!

**earboth** · December 12th 2007, 07:02 AM

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

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

If you have

$g^2 - 6g$ 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 $\left(\frac{-6}{2}\right)^2 = (-3)^2 = 9$ .
Thus

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

**Morzilla** · December 13th 2007, 05:14 AM

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

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

If you have

$g^2 - 6g$ 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 $\left(\frac{-6}{2}\right)^2 = (-3)^2 = 9$ .
Thus

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

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

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