# MX1 Help (1 Viewer)

#### Sxmm

##### New Member
Stuck on these 4 questions, would be great if someone helped out

#### Qeru

##### Well-Known Member
Stuck on these 4 questions, would be great if someone helped out

View attachment 30199
The idea is to complete the square

$\bg_white x^2+4x+4+y^2-2y+1+1=5$

$\bg_white \therefore (x+2)^2+(y-1)^2=4$

Ill let you do the rest (they all following the same idea).

#### YonOra

##### Well-Known Member
Alright so,

Rearragnge the equation to better see what to do! $\bg_white x^2 + 4x + y^2 - 2y = -1$
Now, complete the square for $\bg_white x^2 + 4x$
to: $\bg_white x^2 + 4x + 4 - 4$ Remember, this is done by using $\bg_white \left (\frac{b}{2} \right )^2$
Do the same with $\bg_white y^2 -2y$
You'll get
$\bg_white (x +2)^2 + (y -1)^2 - 5 = -1$
Which is $\bg_white (x +2)^2 + (y -1)^2 = 4$

Do this for the rest.

Yo Latex is a blast. Never put so much effort into the steps of a simple question

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#### Qeru

##### Well-Known Member
$\bg_white (x^2 +2)^2 + (y -1)^2 - 5 = -1$
Which is $\bg_white (x^2 +2)^2 + (y -1)^2 = 4$
I would fix up the typo here