Three real no.x, y, z are such that x^2+6y=-17, y^2 +4z=1 and z^2+2x=2 .
What is the value of x^2 +y^2 +z^2?
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Answers
Given : x² + 6y = -17, y² + 4z = 1 , z² + 2x = 2
To find : x² + y² + z²
Solution:
x² + 6y = -17,
y² + 4z = 1
z² + 2x = 2
Adding all three
x² + y² + z² + 2x + 4z + 6y = -14
=> (x + 1)² - 1 + (y + 3)² -9 + (z + 2)² - 4 = -14
=> (x + 1)² + (y + 3)² + (z + 2)² = 0
as all terms are sqaure andsum zero
hence each term = 0
=> x = - 1 , y = - 3 , z = - 2
x² + y² + z² = 1 + 9 + 4 = 14
x² + y² + z² = 14
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answer is 14 . ok mark me as branliest
