Question

If the ordered triplets of real numbers (x, y, z) satisfy √x-y+z=√x-√y+√z, x+y+z=8 and x-y+z=4, then find the value of xyz.

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Answer 1

Given : √(x - y + z)  = √x-√y+√z , x+y+z=8 and x-y+z=4

To find : value of xyz

Solution:

x+y+z=8

x-y+z=4

=> 2y = 4

=> y = 2

=> x  + z  = 6

√(x - y + z)  = √x-√y+√z

=> √4  =  √x - √2  + √z

=>   √x + √z  =  2  +  √2

Squaring both sides

=> x + z  + 2√x √z   = 4 + 2 + 4√2

=> 6 + 2√x √z   = 6 + 4√2

=> 2√x √z   = 4√2

=> √x √z   =2√2

Squaring both sides

=> xz  = 8

xyz  = (xz)y = 8 * 2   = 16

xz  = 8  , x  + z  = 6

=> x . z  = ( 4 , 2)  or ( 2 , 4 )

x , y , z  are  ( 2 , 2 , 4)  or  ( 4, 2 , 2)

xyz  = 16

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