Find the value of $x^2 + y^2 + z^2$.

Question

If the real numbers $x, y, z$ are such that $x^2 +4y^2 +16z^2=48$ and $xy+4yz+2zx=24$, What is the value 0f $x^2+y^2+z^2 ?$.

The value of $x+2y+4z = \left\lvert 12 \right\rvert$. I don't know how to proceed after that.

Answer 1

$$x^2+4y^2+16z^2=48$$ $$xy+4yz+2zx=24$$ Now, $$2(x^2+4y^2+16z^2)-4(xy+4yz+2zx)=0$$ $$(x-2y)^2+(2y-4z)^2+(x-4z)^2=0$$ Let say $x=2y=4z=k$ $$x=k,y=\frac k2,z=\frac k4$$ Since $k^2+\frac{4k^2}{4}+\frac{16k^2}{16}=48$ $$k=4$$ $$x=4,y=2,z=1$$ $$x^2+y^2+z^2=16+4+1=21$$