Question

if x+y+z=10 and x^2+y^2+z^2=40, then find xy+yz+zx​

Answer 1

$\large{\underline{\underline{\mathfrak{\green{\sf{Solution:-}}}}}}$.

$\large{\underline{\underline{\mathfrak{\pink{\sf{Given\:Here:-}}}}}}$.

✴$\:x+y+z\:=\:10.........(1)$.

✴$\:x^2+y^2+z^2\:=\:40.......(2)$.

$\large{\underline{\underline{\mathfrak{\green{\sf{Find\:Here:-}}}}}}$.

✴$\:Value\:of\:xy+yz+zx$.

$\large{\underline{\underline{\mathfrak{\green{\sf{Explanation:-}}}}}}$.

➡We know that

✴$\:(a+b+c)^2\:=a^2+b^2+c^2+2(ab+bc+ca)$.

➡Squaring both side. of (1) equation .

$\implies\:(x+y+z)^2\:=10^2$.

$\implies\:(x^2+y^2+z^2)\:+\:2(xy+yz+zx)\:=100$.

➡Keep value by (1) And (2).

$\implies\:(40)\:+\:2(xy+yz+zx)\:=\:100$.

$\implies\:2(xy+yz+zx)\:=\:100\:-\:40$.

$\implies\:(xy+yz+zx)\:=\frac{60}{2}$.

$\implies\:(xy+yz+zx)\:=\:30$.

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