# somebody please teach me to complete the square

• Dec 3rd 2006, 06:33 PM
jhonwashington
somebody please teach me to complete the square
hello

could you please be that nice to teach me how to complete the square, step by step, i think i understand most of it, except when it comes to factorize, i'm being taught about ellipses and hyperbolas and i'm having a very hard time because i don't know how to complete the square

for example how would you solve this excercise

4x^2+3y^2+8x-6y=0

=========================
this is whow i would do it

2 (2x^2 +4+4) 3(y^2-2y+2) = 0

ok I give up, i don't know how to do it, please help me
thank you.
• Dec 3rd 2006, 06:52 PM
ThePerfectHacker
Quote:

Originally Posted by jhonwashington

4x^2+3y^2+8x-6y=0

First you need to have the squared coefficient free, that is, equal to 1.
You do this by factoring,
$\displaystyle (4x^2+8x)+(3y^2-6y)=0$
Factor,
$\displaystyle 4(x^2+2x)+3(y^2-6y)=0$
Now look at the linear terms (2 and -6)
Add half the number squared and subtract,
$\displaystyle 4(x^2+2x+1-1)+3(y^2-6y+9-9)=0$
Distribute in the following strange way,
$\displaystyle 4(x^2+2x+1)-4(1)+3(y^2-6y+9)-3(9)=0$
You are about to see why we distribure like that.
Now you should see the perfect squares.
$\displaystyle 4(x+1)^2-4+3(y-3)^2-27=0$
Bring to the other side the free terms,
$\displaystyle 4(x+1)^2+3(y-3)^2=31$
• Dec 3rd 2006, 08:25 PM
earboth
Quote:

Originally Posted by ThePerfectHacker
...
You do this by factoring,
$\displaystyle (4x^2+8x)+(3y^2-6y)=0$
Factor,
$\displaystyle 4(x^2+2x)+3(y^2-6y)=0$
...

Hello TPH,

it looks to me as if you have made a typo here: $\displaystyle 4(x^2+2x)+3(y^2-$6y$\displaystyle )=0$

EB
• Dec 4th 2006, 05:42 AM
Soroban
Hello, jhonwashington!

This problem has particularly ugly numbers . . . I'll modify it.

This is the approach I've taught in my classes.

Quote:

$\displaystyle 4x^2 + 3y^2 + 8x - 6y\:=$ 5

We have: .$\displaystyle 4x^2+8x + 3y^2-6y\:=\:5$ .

Factor "in groups": .$\displaystyle 4(x^2 + 2x\qquad) + 3(y^2 - 2y\qquad) \:=\:5$

This is the complete-the-square step:
. . Take one-half of the coefficient of the linear term and square it.
. . "Add to both sides."

The coefficient of $\displaystyle x$ is $\displaystyle 2.$
. . $\displaystyle \frac{1}{2}(2) = 1\quad\Rightarrow\quad 1^2 = 1$

So we "add to both sides" . . but be careful!

We have: .$\displaystyle 4(x^2 + 2x \,+\,$1$\displaystyle ) + 3(y^2 - 2y\qquad)\:=\:5\,+$4 .
Why 4 ?
. . . . . . . . $\displaystyle \hookrightarrow$ . . . . . $\displaystyle \uparrow$
. . . . . . . . .
We wrote $\displaystyle +1$ on the left side
. . . . . . . .
but it is multiplied by the leading 4.
. . . . . . .
So we actually added 4 to the left side.

Complete the square for the $\displaystyle y$-terms.
. . $\displaystyle \frac{1}{2}(-2) = -1\quad\Rightarrow\quad (-1)^2=1$

"Add to both sides": .$\displaystyle 4(x^2 + 2x + 1) + 3(y^2 + 2y \,+\,$1$\displaystyle ) \;=\;9 \,+$ 3

Factor: .$\displaystyle 4(x+1)^2 + 3(y-1)^2\;=\;12$

Divide by $\displaystyle 12\!:\;\;\frac{4(x+1)^2}{12} + \frac{3(y-1)^2}{12}\;=\;1$

Then we have: .$\displaystyle \frac{(x+1)^2}{3} + \frac{(y-1)^2}{4} \;=\;1$

The ellipse is centered at $\displaystyle (-1,1)$
Its semiminor axis (x- direction) is: $\displaystyle \sqrt{3}$
Its semimajor axis (y-direction) is: $\displaystyle 2$

• Dec 4th 2006, 11:51 AM
jhonwashington
Thank you so much for the help ThePerfectHacker ,earboth and soroban,

just one last question, how do you guys factorize? for example how did

4(x^2+2x+1) becomes
4 (x+1)^2

again thanks a lot for the help
• Dec 4th 2006, 12:00 PM
topsquark
Quote:

Originally Posted by jhonwashington
Thank you so much for the help ThePerfectHacker ,earboth and soroban,

just one last question, how do you guys factorize? for example how did

4(x^2+2x+1) becomes
4 (x+1)^2

again thanks a lot for the help

Note:
$\displaystyle (x + a)^2 = x^2 + 2ax + a^2$

-Dan
• Dec 4th 2006, 12:02 PM
ThePerfectHacker
Quote:

Originally Posted by jhonwashington

4(x^2+2x+1) becomes
4 (x+1)^2

again thanks a lot for the help

Whenever you add the the half term squared Then you can always factor.

For example,
$\displaystyle x^2+10x$
Add subtract half term squared,
$\displaystyle x^2+10x+25-25$
Thus,
$\displaystyle (x+5)^2-25$.
Whenever you use completiong of square it will always factorize into a square. That is why it is called "completing the square".