Question

If, x, y, z are real numbers such that x + y + z = 4 and x2 + y2 + z2 = 6, then x, y, z lie in:​

Answer 1

SOLUTION

TO DETERMINE

If, x, y, z are real numbers such that x + y + z = 4 and x² + y² + z² = 6, then x, y, z lie in

EVALUATION

Here it is given that

x + y + z = 4 - - - - - - - (1)

Again

x² + y² + z² = 6 - - - - - - (2)

From Equation 1 we get

x = 4 - y - z

From Equation 2 we get

$\sf {x}^{2} + {y}^{2} + {z}^{2} = 6$

$\sf \implies {(4 - y - z)}^{2} + {y}^{2} + {z}^{2} = 6$

$\sf \implies {(y + z - 4)}^{2} + {y}^{2} + {z}^{2} = 6$

$\sf \implies {y}^{2} + {z}^{2} + 16+ 2yz - 8y - 8z + {y}^{2} + {z}^{2} = 6$

$\sf \implies 2 {y}^{2} + 2 {z}^{2} + 2yz - 8y - 8z + 10 = 0$

$\sf \implies {y}^{2} + {z}^{2} + yz - 4y - 4z + 5 = 0$

$\sf \implies {y}^{2} + y(z - 4) + ({z}^{2} - 4z + 5) = 0$

This is an quadratic equation in y

Since y is real

∴ Discriminant ≥ 0

$\displaystyle \sf{ \implies {(z - 4)}^{2} - 4.1.( {z}^{2} - 4z + 5) \geqslant 0}$

$\displaystyle \sf{ \implies {z}^{2} - 8z + 16 - 4{z}^{2} + 16z -20 \geqslant 0}$

$\displaystyle \sf{ \implies - 3{z}^{2} + 8z - 4 \geqslant 0}$

$\displaystyle \sf{ \implies 3{z}^{2} - 8z + 4 \leqslant 0}$

$\displaystyle \sf{ \implies 3{z}^{2} - (6 + 2)z + 4 \leqslant 0}$

$\displaystyle \sf{ \implies 3{z}^{2} - 6z - 2z + 4 \leqslant 0}$

$\displaystyle \sf{ \implies 3z(z - 2) - 2(z - 2) \leqslant 0}$

$\displaystyle \sf{ \implies (3z - 2) (z - 2) \leqslant 0}$

$\displaystyle \sf{ \implies z \in \: \bigg[ \frac{2}{3} ,2 \bigg] \: }$

Similarly it can be shown that

$\displaystyle \sf{ x\in \: \bigg[ \frac{2}{3} ,2 \bigg] \: \: and \: \: y\in \: \bigg[ \frac{2}{3} ,2 \bigg] }$

FINAL ANSWER

Hence x , y , z lies in the interval

$\displaystyle \sf{ \bigg[ \frac{2}{3} ,2 \bigg] }$

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