I need help with this question: Find all the x-coordinates of the points on the curve x^2y^2+xy=2 where the slope of the tangent line is −1.

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I need help with this question: Find all the x-coordinates of the points on the curve x^2y^2+xy=2 where the slope of the tangent line is −1.

Mathematics
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I've used implicit differentiation to get \[2xy^2+x^22yy'+y+xy=0\] I then substituted y' with -1 and got \[2xy^2-x^22y+y-x\]
I don't know what to do next...
so you have \[2xy^2+x^22yy'+y+xy'=0 \\ \text{ okay and you replaced } y' \text{ with } -1 \\ 2xy^2-2x^2y+y-x=0 \\ 2xy(y-x)+1(y-x)=0 \\ (y-x)(2xy+1)=0 \\ \text{ so we have } y=x \text{ or } xy=\frac{-1}{2} \\ \text{ so try plugging both of these possibilities back in the original equation }\]

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So wherever I have a y I replace with x, and wherever I have xy I replace with -1/2 ?
yeah we will see one equation is never true
and you will see one equation is true for 2 real values
2 real x values
for example with the xy=-1/2 sub \[(xy)^2=(\frac{-1}{2})^2 \implies x^2y^2=\frac{1}{4} \\ \text{ but } \frac{1}{4}+\frac{-1}{2} \text{ is never } 2 \]
you can solve the one with y=x pretty easily
Hmmm... but replacing y with x I get ... x^4 + x^2 = 2
right that is a quadratic in terms of x^2
do you know how to solve u^2+u=2?
Yes! Oh so now I see it
I replaced x^2 with u so we can solve for u then replace u with x^2 and then solve for x
Ah thank you so much @freckles !!! I got -1, and 1 and it's correct!
great! :)
I appreciate the help! ^.^
np

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