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.
$\displaystyle y^2+y(3x+1)+(10x^2+5x)=0$Originally Posted by a4swe
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.
Those cannot be the answers.Originally Posted by a4swe
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.
Ahh yes, I just read this now.Originally Posted by a4swe
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.
Hello, a4swe!
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: .$\displaystyle \left\{\begin{array}{cc}y + 5x\:=\:0\quad\Rightarrow\quad \boxed{y \,= \,-5x} \\ y - 2x + 1\:=\:0\quad\Rightarrow\quad \boxed{y\:=\:2x - 1}\end{array}$