If x + y + z =10 and xy + yz + zx = 24, find the value of x2
+ y2
+ z2
.
Answers
Answer:
x+y+z=10 xy+yz +zx=24
Step-by-step explanation:
(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)
(10)^2=x^2+y^2+z^2+2(24)
100=x^2+y^2+z^2+48
100-48=x^2+y^2+z^2
52=x^2+y^2+z^2
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Answer:
138
Step-by-step explanation:
x + y + z = 10
(x + y + z)² = (10)²
x² + y² + z² + 2xy + 2yz + 2zx = 100
x² + y² + z² + 2(xy + yz + zx) = 100 { xy+yz+zx=31}
x² + y² + z² + 2(31) = 100
x² + y² + z² + 62 = 100
x² + y² + z² = 100 - 62
x² + y² + z² = 38 -------{i}
(x+y)² = x² + y² + 2xy
(y+z)² = z² + y² + 2zy
(z+x)² = z² + x² + 2zx
(x+y)² + (y+z)² + (z+x)²
= (x² + y² + 2xy) + (z² + y² + 2zy) + (z² + x² + 2zx)
= x² + y² + z² + y² + z² + x² + 2xy + 2zy + 2zx
= 2x² + 2y² + 2z²
= 2 (x² + y² + z² + xy + zy + zx) -------{ii}
from {i} and {ii}
= 2 (x² + y² + z² + xy + zy + zx) { xy+yz+zx=31}
= 2 (38 + 31)
= 2 (69)
= 138
∴(x+y)² + (y+z)² + (z+x)² = 138