19. Shan solves the simultaneous equations
$$
x y=15 \text { and }(2 x-y)^{4}=1
$$
where $x$ and $y$ are real numbers. She calculates $z$, the sum of the squares of all the $y$-values in her solutions.
What is the value of $z$ ?
SolUtion
122
From $(2 x-y)^{4}=1$ we know $(2 x-y)^{2}= \pm 1$. Since any squared quantity must be non-negative, we know $(2 x-y)^{2}=1$ from which $2 x-y= \pm 1$.
Consider the case $2 x-y=1$. Multiplying by $y$ gives $2 x y-y^{2}=y$, but $x y=15$ and so $30-y^{2}=y$. Therefore $y=-6,5$.
Consider the case $2 x-y=-1$. Multiplying by $y$ gives $2 x y-y^{2}=-y$, but $x y=15$ and so $30-y^{2}=-y$. Therefore $y=-5,6$.
Hence $z=(-6)^{2}+5^{2}+(-5)^{2}+6^{2}=122$.