14. The graph of $\left(x^{2}+y^{2}-2 x\right)^{2}=2\left(x^{2}+y^{2}\right)^{2}$ meets the $x$-axis in $p$ different places and meets the $y$-axis in $q$ different places.
What is the value of $100 p+100 q$ ?
SolUtion
400

The graph of $\left(x^{2}+y^{2}-2 x\right)^{2}=2\left(x^{2}+y^{2}\right)^{2}$ meets the $x$-axis when $y=0$.
i.e. $\left(x^{2}-2 x\right)^{2}=2\left(x^{2}\right)^{2}$
i.e. $x^{4}-4 x^{3}+4 x^{2}=2 x^{4}$
i.e. $0=x^{4}+4 x^{3}-4 x^{2}$
i.e. $x=0$ or $0=x^{2}+4 x-4$
i.e. $x=0$ or $x=-2 \pm 2 \sqrt{2}$.

Therefore there are three points where the graph meets the $x$-axis, so $p=3$.
The graph of $\left(x^{2}+y^{2}-2 x\right)^{2}=2\left(x^{2}+y^{2}\right)^{2}$ meets the $y$-axis when $x=0$, i.e. $\left(y^{2}\right)^{2}=2\left(y^{2}\right)^{2}$
i.e. $y^{4}=2 y^{4}$, with only one solution of $y=0$.

Therefore there is only one point where the graph meets the $y$-axis, so $q=1$.
The value of $100 p+100 q$ is $100 \times 3+100 \times 1=400$.