Let `x ,y ,z in R` such that `x_y+z=6a n xx y+y z+z x=7.` Then find the range of values of `x ,y ,a n dzdot`

Answer

` a in (- infty, - 1//2 )sup [ 4, nfty)`

Explanation

x,y,z,`in` R
` x + y + z = 6 and xy + yz + zx = 7`
`rArr y (6 - y - z) + yz + z (6 - y - z) = 7`
`rArr - y^(2) + (6 - z + z - z) y + z (6 -z) - 7 = 0`
`rArr y^(2) + (z - 6 )y + 7 + z (z - 6) = 0`
Now, y is real. Therefore ,
`(z - 6)^(2) - 4 [7 + z (z - 6)] ge 0 `
or ` 3z^(2) - 12z - 8 le 0`
or `(12 - sqrt(144 + 96))/(6) le zle (12 + sqrt(144 + 96))/(6)`
or ` (6 - 2 sqrt(15))/(3) le z le (6 + 2 sqrt(15))/(3)`.
From symmetry, x and y have same range.