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

  • Thread starter greko
  • Start date
The equation can be rewritten as (y-x)^2 = -2x + 12y + 10, which is equivalent to 4y^2 - 4xy + x^2 + (4x-12)y + (x^2 - 10) = 0. So, a=4, b=4x-12, c=x^2-10.
  • #1
greko
8
0
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.
 
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  • #2
greko said:
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.
 
  • #3
Even so I can't solve for y. Can you put a step by step explanation?
 
  • #4
The closest I get is (y-x)^2-3y=(3x^2-x+10)/4, how can i proceed?
 
  • #5
It is forbidden on these forums to post step-by-step solutions to a posted problem.
 
  • #6
sorry, I am new, but can you explain what do I do after, (y-x)^2-3y=(3x^2-x+10)/4?
 
  • #7
Start by expanding (y-x)^2...then collect terms in powers of y.
 
  • #8
gabbagabbahey said:
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.
 
  • #9
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.
 
  • #10
gabbagabbahey said:
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.
 
  • #11
yup.
 
  • #12
greko said:
The closest I get is (y-x)^2-3y=(3x^2-x+10)/4, how can i proceed?

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

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

greko said:
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?
 
  • #13
The Chaz said:
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.
 
  • #14
Cool man. Hope to help more (or some :wink:) in the future!
 
  • #15
The Chaz said:
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...
 

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

What is a quadratic equation?

A quadratic equation is a mathematical equation of the form ax^2 + bx + c = 0, where a, b, and c are constants and x is the variable. It is called a quadratic equation because the highest power of the variable is 2.

How do you solve a quadratic equation?

To solve a quadratic equation, you can use the quadratic formula x = (-b ± √(b^2 - 4ac)) / 2a, where a, b, and c are the constants in the equation. You can also factor the equation or use completing the square method to solve it.

What is the discriminant of a quadratic equation?

The discriminant of a quadratic equation is the expression b^2 - 4ac. It is used to determine the nature of the roots of the equation. If the discriminant is positive, there are two distinct real roots. If it is zero, there is one real root. And if it is negative, there are no real roots.

How do you know if a quadratic equation has real or complex roots?

A quadratic equation has real roots if the discriminant b^2 - 4ac is greater than or equal to 0. If the discriminant is negative, then the equation has complex roots.

How do you solve the quadratic equation x^2-4xy+4y^2+x-12y-10=0?

To solve the given quadratic equation, we can first rearrange it in the standard form ax^2 + bx + c = 0. In this case, it becomes x^2 - 4xy + 4y^2 + x - 12y - 10 = 0. Then, we can factor the quadratic expression (x - 2y)^2 and simplify it to (x - 2y)^2 + (x - 2y) - 10 = 0. Using the quadratic formula, we can get the two solutions x = 2y + 5 and x = 2y - 1.

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