find x^2+y^2+z^2 if x^2+xy+xz=135,y^2+yz+yx=351 and z^2+zx+zy=243.
Answers
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Please find below the solution to the asked query:
Given : x2 + xy + xz = 135 --- ( 1 ) ,
y2 + yz + yx = 351 --- ( 2 )
And
z2 + zx + zy = 243 --- ( 3 )
Now we add equation 1 , 2 and 3 and get
x2 + y2 + z2 + 2 xy + 2 yz + 2 zx = 729
( x + y + z )2 = 729 ( We know : ( x + y + z )2 = x2 + y2 + z2 + 2 xy + 2 yz + 2 zx )
( x + y + z )2 = 272
x + y + z = 27 --- ( 4 )
Now we take equation 1 : x2 + xy + xz = 135
x ( x + y + z ) = 135 , Substitute value from equation 4 and get
27 x = 135
x = 5
And we take equation 2 : y2 + yz + yx = 351
y ( x + y + z ) = 351 , Substitute value from equation 4 and get
27 y = 351
y = 13
And we take equation 3 : z2 + zx + zy = 243
z ( x + y + z ) = 243 , Substitute value from equation 4 and get
27 z = 243
z = 9
Therefore,
Value of x2 + y2 +z2 = 52 + 132 + 92 = 25 + 169 + 81 = 275 ( Ans )
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Answer:
275
Step-by-step explanation:
Given : x2 + xy + xz = 135 --- ( 1 ) ,
y2 + yz + yx = 351 --- ( 2 )
And
z2 + zx + zy = 243 --- ( 3 )
Now we add equation 1 , 2 and 3 and get
x2 + y2 + z2 + 2 xy + 2 yz + 2 zx = 729
( x + y + z )2 = 729 ( We know : ( x + y + z )2 = x2 + y2 + z2 + 2 xy + 2 yz + 2 zx )
( x + y + z )2 = 272
x + y + z = 27 --- ( 4 )
Now we take equation 1 : x2 + xy + xz = 135
x ( x + y + z ) = 135 , Substitute value from equation 4 and get
27 x = 135
x = 5
And we take equation 2 : y2 + yz + yx = 351
y ( x + y + z ) = 351 , Substitute value from equation 4 and get
27 y = 351
y = 13
And we take equation 3 : z2 + zx + zy = 243
z ( x + y + z ) = 243 , Substitute value from equation 4 and get
27 z = 243
z = 9
Therefore,
Value of x2 + y2 +z2 = 52 + 132 + 92 = 25 + 169 + 81 =275
Thanks
Aarohi