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One of the questions on my Algebra 1 homework was to find all values of $y$ such that $$(y^2+y-6)(x^2-6y+9)-2(y^2-9) = 0$$

This was listed under "Lecture 13: Quadratic Equations"

I can not figure out how to approach this problem? Should I expand it? I also noticed that one of the terms used $x$.

J_P
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Jamlandia
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  • That's what it says in the textbook, although it could be a typo – Jamlandia Jun 01 '19 at 11:04
  • Even if $x$ is not a typo, you can extract a fairly obvious factor $y+3$ noting the solution $y=-3$ independent of $x$ and you are then left with a quadratic in $y$ to solve. – Mark Bennet Jun 01 '19 at 11:26
  • It is very likely to be a typo. I checked the equation using WolframAlpha. There are $3$ lines in the graph, namely $y = -3$, $ y = 1/12 (x^2 - \sqrt{x^4 - 10 x^2 + 73} + 19)$ and $y = 1/12 (x^2 + \sqrt{x^4 - 10 x^2 + 73} + 19)$ – Y.T. Jun 01 '19 at 14:09

2 Answers2

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Write it like this $$(y-2)(y+3)(y-3)^2+2(y-3)(y+3)=0$$

now $$(y-3)(y+3)\cdot \Big((y-2)(y-3)+2\Big)=0$$

Can you do it now?

nonuser
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$$(y^2+y-6)(x^2-6y+9)-2(y^2-9) = 0 \\ (y-2)(y+3)(x^2-6y+9)-2(y-3)(y+3)=0 \\ (y-2)(y+3)(x^2-6y+9)=2(y-3)(y+3) \\ (y-2)(x^2-6y+9)=2(y-3) \\ x^2-6y+9=\frac{2(y-3)}{y-2} \\ x^2=\frac{2(y-3)}{y-2}+6y-9 \\ x^2= \frac{2(y-3)}{y-2}+\frac{6y(y-2)}{y-2}-\frac{9(y-2)}{y-2} \\x^2= \frac{2(y-3)+6y(y-2)-9(y-2)}{y-2} \\ x^2=\frac{2y-6+6y^2-12y-9y+18}{y-2}=\frac{6y^2-19y+12}{y-2} \\ x^2=\frac{6y^2-19y+12}{y-2}$$ That's the simplest form for the equation you've written above; However there isn't any unique root (solution) for the equation; as the equation has two variables and needs two equations to find unique $(x,y)$, so solving for $x$ and $y$ can't be; because there isn't enough given. The equation above is said to be a "Relation between $x$ and $y$" Unless the $x$ in the equation become $y$.

Graph Graphed using: Geogebra.

Anas Khaled
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