Question:
Find the value of the equation: $4 x^{2}+y^{2}+25 z^{2}+4 x y-10 y z-20 z x$ when $x=4, y=3, z=2$
Solution:
$4 x^{2}+y^{2}+25 z^{2}+4 x y-10 y z-20 z x$
$(2 x)^{2}+y^{2}+(-5 z)^{2}+2(2 x)(y)+2(y)(-5 z)+2(-5 z)(2 x)$
$(2 x+y-5 z)^{2}$
$(2(4)+3-5(2))^{2}$
$(8+3-10)^{2}$
$(1)^{2}$
1
Hence value of the equation is equals to 1