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.
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How did you find the value of $x+2y+4z$? And what is $|12|$ supposed to signify? Isn't that just $12$? – Arthur Jun 28 '18 at 15:29
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How about solving $x$, $y$, $z$ from what you've given and then plugging them to either of the two equations given or into $x^2+y^2+z^2$? I'd expect the general way to study these things to be to first try to simplify the equations that you've been given. – mavavilj Jun 28 '18 at 15:33
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1@Arthur $(x+2y+4z)^2$ can be expanded out to $x^2+4y^2+16z^2+4(xy+2xz+4yz)$, which is equal to 48+4*24=144. So $(x+2y+4z)^2=144$. – Acccumulation Jun 28 '18 at 15:37
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$$x^2+4y^2+16z^2=48$$ $$xy+4yz+2zx=24$$ 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 $k^2+\frac{4k^2}{4}+\frac{16k^2}{16}=48$ $$k=4$$ $$x=4,y=2,z=1$$ $$x^2+y^2+z^2=16+4+1=21$$
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1@egreg true, but I think it should be said. The answer is a good one. – Mark Bennet Jun 28 '18 at 16:32