# y^2+3xy-10x^2+y+5x=0 solving guide lines

• Aug 24th 2006, 02:37 PM
a4swe
y^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.
• Aug 24th 2006, 02:45 PM
ThePerfectHacker
Quote:

Originally Posted by a4swe
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.

$\displaystyle y^2+y(3x+1)+(10x^2+5x)=0$
Thus,
$\displaystyle y=\frac{-3x-1\pm \sqrt{(3x+1)^2-4(10x^2+5x)}}{2}$
Thus,
$\displaystyle y=\frac{-3x-1\pm \sqrt{9x^2+6x+1-40x^2-20x}}{2}$
Thus,
$\displaystyle y=\frac{-3x-1\pm \sqrt{-31x^2-12x}}{2}$
Exactly, what are you trying to do?
Graph the curve?
If thus, it is a conic and there are simpler ways.
• Aug 24th 2006, 02:50 PM
a4swe
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.
• Aug 24th 2006, 02:53 PM
a4swe
And hrmmm, I got it wrong typing the keyboard the second time to, it should be -10x^2.
• Aug 24th 2006, 04:16 PM
Soltras
Quote:

Originally Posted by a4swe
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.

Want to solve $\displaystyle y^2+3xy+10x^2+y+5x=0$.

Supposing $\displaystyle y=2x-1$ were a solution then:

$\displaystyle (2x-1)^2+3x(2x-1)+10x^2+(2x-1)+5x=0$
$\displaystyle (4x^2-4x+1)+(6x^2-3x)+10x^2+2x-1+5x = 0$
$\displaystyle 20x^2 = 0$

This isn't generally true (it is only when $\displaystyle x=0$, but that hasn't been stipulated.)

Now, supposing the other possibility, then $\displaystyle y=-5x$:

$\displaystyle (-5x)^2+3x(-5x)+10x^2+(-5x)+5x = 0$
$\displaystyle 25x^2-15x^2+10x^2=0$
$\displaystyle 20x^2 = 0$

Again, interestingly we are left with this supposed equality of zero and $\displaystyle 20x^2$, which isn't true.

Since your textbook identifies those as answers, is it possible you left a $\displaystyle -20x^2$ 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.
• Aug 24th 2006, 04:20 PM
Soltras
Quote:

Originally Posted by a4swe
And hrmmm, I got it wrong typing the keyboard the second time to, it should be -10x^2.

Ahh yes, I just read this now.
Going from $\displaystyle +10x^2$ to $\displaystyle -10x^2$ is a difference of . . . $\displaystyle 20x^2$, 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.
• Aug 24th 2006, 09:45 PM
Soroban
Hello, a4swe!

Quote:

Solve for $\displaystyle y:\;\;y^2+3xy-10x^2+y+5x\;=\;0$

The expression factors!

We have: .$\displaystyle (y^2 + 3xy - 10x^2) + (y + 5x)\;=\;0$

Factor: .$\displaystyle (y - 2x)\underbrace{(y + 5x)} + 1\underbrace{(y + 5x)}\;=\;0$

Factor: .$\displaystyle (y + 5x)(y - 2x + 1)\;=\;0$

Therefore, we have: .$\displaystyle \left\{\begin{array}{cc}y + 5x\:=\:0\quad\Rightarrow\quad \boxed{y \,= \,-5x} \\ y - 2x + 1\:=\:0\quad\Rightarrow\quad \boxed{y\:=\:2x - 1}\end{array}$

• Aug 25th 2006, 04:27 AM
a4swe
Nice!
I solved it using the "ThePerfectHacker method" first, but the second one was easier.
Thank you both, I can now sleep tight again.