# To see a solution to a 2nd degree equation without using the regular formula

• Jul 30th 2011, 12:01 PM
fysikbengt
To see a solution to a 2nd degree equation without using the regular formula
This might be a little to advanced for this forum, but I think it is a little bit to easy to post in linear algebra. My question is if there is a quick way to see the solution of this equation, solving y as a function of x, without using the 2nd degree equation formula. Here it is:
\$\displaystyle y^{2}(1+4a)-8axy+x^{2}(-1+4a)=0\$
• Jul 30th 2011, 12:25 PM
earboth
Re: To see a solution to a 2nd degree equation without using the regular formula
Quote:

Originally Posted by fysikbengt
This might be a little to advanced for this forum, but I think it is a little bit to easy to post in linear algebra. My question is if there is a quick way to see the solution of this equation, solving y as a function of x, without using the 2nd degree equation formula. Here it is:
\$\displaystyle y^{2}(1+4a)-8axy+x^{2}(-1+4a)=0\$

1. By first inspection you see that y = x.

2. Divide the LHS of the equation by (y - x). (Use long division)
You should come out with

x·(1 - 4·a) + y·(4·a + 1)

3. Now solve for y:

x·(1 - 4·a) + y·(4·a + 1) = 0
• Jul 30th 2011, 03:20 PM
fysikbengt
Re: To see a solution to a 2nd degree equation without using the regular formula
Quote:

Originally Posted by earboth
1. By first inspection you see that y = x.

2. Divide the LHS of the equation by (y - x). (Use long division)
You should come out with

x·(1 - 4·a) + y·(4·a + 1)

3. Now solve for y:

x·(1 - 4·a) + y·(4·a + 1) = 0

Thanks, I dont think I would have come up with that myself, but now it is clear. From the physical interpretation it is obvious that y = x is a (superfluos) solution, and although it is not trivial to perform the long division it is easier if you know what you are looking for.