MHB Solving a Quadratic Equation with Two Variables

[Amer]

Solve for $x$:

$$4x^2 +4xy - 6y - 9 = 9 $$

 

[MarkFL]

I would write the equation as a quadratic in x in standard form:

$\displaystyle 4x^2+4yx-(6(y+3))=0$

Now apply the quadratic formula.

 

[Amer]

Completing the square

$ (2x)^2 + 2(2x)y -6y - 9 = 0 $
$ (2x+y)^2 -y^2 -6y - 9 = 0 $
$(2x+y)^2 - (y+3)^2 = 0 $
$(2x+y)^2 = (y+3)^2 $
$ 2x+y = |y+3| $

 

[MarkFL]

What happened to the 9 originally on the right side?

 

[Amer]

What happened to the 9 originally on the right side?

lol typo

 

[MarkFL]

I've made a few of those in my time! (Tauri)

 

[CaptainBlack]

Completing the square

$ (2x)^2 + 2(2x)y -6y - 9 = 0 $
$ (2x+y)^2 -y^2 -6y - 9 = 0 $
$(2x+y)^2 - (y+3)^2 = 0 $
$(2x+y)^2 = (y+3)^2 $
$ 2x+y = |y+3| $

Last line should be:

\(2x+y = \pm(y+3)\)

CB

 

[Amer]

Last line should be:

\(2x+y = \pm(y+3)\)

CB

what is wrong in my solution ?

 

[earboth]

Solve for x:

$$4x^2 +4xy - 6y - 9 = 9 $$

what is wrong in my solution ?

Since x is the variable and y is considered to be a constant your solution should read

$$|2x+y| = y+3$$

as CB has pointed out.

 

[freiermuthj]

Or you can use POLYSMLT if you have a TI83