Solve Quadratic Equation x^2-4xy+4y^2+x-12y-10=0

[greko]

1. Solve for y explicitly , x^2-4xy+4y^2+x-12y-10=0

Homework Equations

3. I reduced it to (x-2y)^2=-2x+12y+10 But I have no idea how to continue.

 

[Mark44]

1. Solve for y explicitly , x^2-4xy+4y^2+x-12y-10=0

Homework Equations

3. I reduced it to (x-2y)^2=-2x+12y+10 But I have no idea how to continue.

Your equation is quadratic in y. Rewrite it as 4y^2 + (stuff)y + (other stuff) = 0, and use the quadratic formula.

 

[greko]

Even so I can't solve for y. Can you put a step by step explanation?

 

[greko]

The closest I get is (y-x)^2-3y=(3x^2-x+10)/4, how can i proceed?

 

[Dickfore]

It is forbidden on these forums to post step-by-step solutions to a posted problem.

 

[greko]

sorry, I am new, but can you explain what do I do after, (y-x)^2-3y=(3x^2-x+10)/4?

 

[gabbagabbahey]

Start by expanding (y-x)^2...then collect terms in powers of y.

 

[greko]

Start by expanding (y-x)^2...then collect terms in powers of y.

From there, I get y^2-2xy-x^2-3y=(3x^2-x+10)/4, I guess I could complete the square on the left side but I wouldn't really get anywhere I think.

 

[gabbagabbahey]

When I say collect terms in powers of y, I mean write -2xy-3y as -(2x+3)y...so you have y^2-(2x+3)y-x^2=(3x^2-x+10)/4...subtract (3x^2-x+10)/4 from both sides of your equation and you can then use the quadratic equation to solve for y in terms of x.

 

[greko]

When I say collect terms in powers of y, I mean write -2xy-3y as -(2x+3)y...so you have y^2-(2x+3)y-x^2=(3x^2-x+10)/4...subtract (3x^2-x+10)/4 from both sides of your equation and you can then use the quadratic equation to solve for y in terms of x.

So would my -x^2-((3x^2-x+10)/4) be considered as my "C" value in my quadratic.

 

[gabbagabbahey]

yup.

 

[The Chaz]

The closest I get is (y-x)^2-3y=(3x^2-x+10)/4, how can i proceed?

sorry, I am new, but can you explain what do I do after, (y-x)^2-3y=(3x^2-x+10)/4?

Start by expanding (y-x)^2...then collect terms in powers of y.

From there, I get y^2-2xy-x^2-3y=(3x^2-x+10)/4, I guess I could complete the square on the left side but I wouldn't really get anywhere I think.

So... why wouldn't we just start with:
x^2-4xy+4y^2+x-12y-10=0
(4)y^2 + (-4x-12)y +(x^2 + x -10) = 0
a, b, c...
instead of completing - then UNcompleting - the square?

 

[greko]

So... why wouldn't we just start with:
x^2-4xy+4y^2+x-12y-10=0
(4)y^2 + (-4x-12)y +(x^2 + x -10) = 0
a, b, c...
instead of completing - then UNcompleting - the square?

Lol, Yes i have noticed that and yes it works!. Thanks for your help guys. I just went back to my original equation and did quadratic formula =D.

 

[The Chaz]

Cool man. Hope to help more (or some ) in the future!

 

[Mark44]

So... why wouldn't we just start with:
x^2-4xy+4y^2+x-12y-10=0
(4)y^2 + (-4x-12)y +(x^2 + x -10) = 0
a, b, c...
instead of completing - then UNcompleting - the square?

Which is what I was talking about in post #2...