y^2+3xy+10x^2+y+5x=0

Any one got any tips on how to solve that one in y?

It has been buging me.

Printable View

- August 24th 2006, 02:37 PMa4swey^2+3xy-10x^2+y+5x=0 solving guide lines
y^2+3xy+10x^2+y+5x=0

Any one got any tips on how to solve that one in y?

It has been buging me. - August 24th 2006, 02:45 PMThePerfectHackerQuote:

Originally Posted by**a4swe**

Thus,

Thus,

Thus,

Exactly, what are you trying to do?

Graph the curve?

If thus, it is a conic and there are simpler ways. - August 24th 2006, 02:50 PMa4swe
I am just trying to get my answer to be equal to the one stated in the textbook, nothing more nothing less.

These answers are by the way y=2x-1 y=-5x and I still don't see any way to get that. - August 24th 2006, 02:53 PMa4swe
And hrmmm, I got it wrong typing the keyboard the second time to, it should be -10x^2.

- August 24th 2006, 04:16 PMSoltrasQuote:

Originally Posted by**a4swe**

Want to solve .

Supposing were a solution then:

This isn't generally true (it is only when , but that hasn't been stipulated.)

Now, supposing the other possibility, then :

Again, interestingly we are left with this supposed equality of zero and , which isn't true.

Since your textbook identifies those as answers, is it possible you left a term off of copying the original question? If not, then your book is wrong and you should go with Hacker's solution, which is correct. - August 24th 2006, 04:20 PMSoltrasQuote:

Originally Posted by**a4swe**

Going from to is a difference of . . . , exactly the term that wouldn't disappear from my work above.

Fix your equation so it's correct and then apply Hacker's technique of grouping it according to parameters for the quadratic formula. - August 24th 2006, 09:45 PMSoroban
Hello, a4swe!

Quote:

Solve for

The expression factors!

We have: .

Factor: .

Factor: .

Therefore, we have: .

- August 25th 2006, 04:27 AMa4swe
Nice!

I solved it using the "ThePerfectHacker method" first, but the second one was easier.

Thank you both, I can now sleep tight again.