Question

If x + y + z = 6, xy + yz + zx = 11. Find the value of x2 + y2 + z2

Answer 1

(x + y + z) = 6 

Square on both sides,

(x + y + z)^2 = 6^2 

x^2 + y^2 + z^2 + 2(xy + yz + zx) = 36 

x^2 + y^2 + z^2 + 2(11) = 36 

x^2 + y^2 + z^2 + 22 = 36 

x^2 + y^2 + z^2 =36 - 22 

x^2 + y^2 + z^2 = 14 



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Answer 2

Heya ✋

Let see your answer !!!!!

Given that

$x + y + z = 6$
$xy + yz + zx = 11$
$x ^{2} + y ^{2} + z ^{2} =$

Solution

$x + y + z = 6 \\ = > on \: squaring \: both \: sides \\ = > (x + y + z) ^{2} = (6) ^{2} \\ = > x ^{2} + y ^{2} + z ^{2} + 2xy + \\ 2yz + 2zx = 36 \\ = > x ^{2} + y ^{2} + z ^{2} + 2(xy + \\ yz + zx) = 36 \\ = > x ^{2} + y ^{2} + z ^{2} + 2 \times 11 = \\ 36 \\ = > x ^{2} + y ^{2} + z ^{2} + 22 = 36 \\ = > x ^{2} + y ^{2} + z ^{2} = 36 - 22 \\ = > x ^{2} + y ^{2} + z ^{2} = 14$





Thanks :)))))