Question

If x square +y square + z square =29 and xy+yz+zx=26 then the value of x+y+z is?

Answer 1

Answer:

formula

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

we know that x^2+y^2+z^2=29

xy+yz+zz=26

so

(x+y+z) ^2=29+2(26)

=29+52

(x+y+z) ^2=81

x+y+z=9

Step-by-step explanation:

answer is 9

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Answer 2

Answer:

If  $x^{2} +y^{2}+z^{2}=29$

and $xy+yz+xz=26$ ,

then the value of   $x+y+z=9$.

Step-by-step explanation:

We know the expansion of

$(x+y+z)^{2} = x^{2} +y^{2}+z^{2} +2xy+2yz+2xz$

                  $=x^{2} +y^{2}+z^{2}+2(xy+yz+xz)$

and it is given that  $x^{2} +y^{2}+z^{2}=29$  and $xy+yz+xz=26$

therefore, $(x+y+z)^{2}=29+2 (26)$  (substituting the values in the expansion)

                                    $=29+52$

                                    $=81$

hence $x+y+z=9$   (since $9^{2}=81$)

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