Question

(15 points) If x, y and z are real numbers such that x^2+y^2+2z^2=4x-2z+2yz-5, then the possible value of (x-y-z) is​

Answer 1

SOLUTION

GIVEN

If x, y and z are real numbers such that

$\sf{ {x}^{2} + {y}^{2} + 2 {z}^{2} = 4x - 2z + 2yz - 5 }$

TO DETERMINE

The possible value of (x-y-z)

EVALUATION

$\sf{ {x}^{2} + {y}^{2} + 2 {z}^{2} = 4x - 2z + 2yz - 5 }$

$\sf{ \implies {x}^{2} - 4x + 4+ {y}^{2} - 2yz+ {z}^{2} + {z}^{2} + 2z +1 = 0 }$

$\sf{ \implies {(x - 2)}^{2} + {(y - z)}^{2} + {(z + 1)}^{2} = 0 }$

We know that if sum of the squares of three Real Numbers are zero then they are separately zero

$\sf{ \implies {(x - 2)} = 0 \: , \: {(y - z)} = 0 \:, \: {(z + 1)} = 0 }$

$\sf{ \implies x = 2 \: , \: y = z \:, \: z = - 1 }$

$\sf{ \implies x = 2 \: , \: y = - 1 \:, \: z = - 1 }$

Hence

$\sf{ x - y - z}$

$= 2 + 1 + 1$

$= 4$

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