Question

If x +y+ z =9 and x y +y z +x z=26 find the value of x2+y2+z2

Answer 1

Answer:

Step-by-step explanation:

Answer 2

If x +y+ z =9 and x y +y z +x z=26 then $\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}=29$

Solution:

Given that x + y + z = 9 , xy + yz + zx = 26

We need to find value of $\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}$

We will be using following algebraic identity

$(\mathrm{x}+\mathrm{y}+\mathrm{z})_{}^{2}=\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}+2(\mathrm{xy}+\mathrm{yz}+\mathrm{zx})$

On substituting given values in above identity we get

$\begin{array}{l}{9^{2}=x^{2}+y^{2}+z^{2}+2(26)} \\\\ {=>81=x^{2}+y^{2}+z^{2}+52} \\\\ {=>x^{2}+y^{2}+z^{2}=81-52=29}\end{array}$

Hence required value of $\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}=29$