# Math Help - somebody please teach me to complete the square

1. ## 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.

2. 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,
$(4x^2+8x)+(3y^2-6y)=0$
Factor,
$4(x^2+2x)+3(y^2-6y)=0$
Now look at the linear terms (2 and -6)
Add half the number squared and subtract,
$4(x^2+2x+1-1)+3(y^2-6y+9-9)=0$
Distribute in the following strange way,
$4(x^2+2x+1)-4(1)+3(y^2-6y+9)-3(9)=0$
You are about to see why we distribure like that.
Now you should see the perfect squares.
$4(x+1)^2-4+3(y-3)^2-27=0$
Bring to the other side the free terms,
$4(x+1)^2+3(y-3)^2=31$

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

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

EB

4. Hello, jhonwashington!

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

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

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

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

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

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

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

We have: . $4(x^2 + 2x \,+\,$1 $) + 3(y^2 - 2y\qquad)\:=\:5\,+$4 .
Why 4 ?
. . . . . . . . $\hookrightarrow$ . . . . . $\uparrow$
. . . . . . . . .
We wrote $+1$ 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 $y$-terms.
. . $\frac{1}{2}(-2) = -1\quad\Rightarrow\quad (-1)^2=1$

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

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

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

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

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

5. 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

6. 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:
$(x + a)^2 = x^2 + 2ax + a^2$

-Dan

7. 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,
$x^2+10x$
Add subtract half term squared,
$x^2+10x+25-25$
Thus,
$(x+5)^2-25$.
Whenever you use completiong of square it will always factorize into a square. That is why it is called "completing the square".