Question 7071
 │x^2 + 5│ + │x - 2│ ≤ 8  ...(**)
 You did wrong at very first step,because you did not apply the def. of abs
 value correctly.

 Note that x^2 + 5 is always positive, so  |x^2 + 5| = x^2 + 5.

 Case (i) if x - 2 >= 0, then |x-2| = x-2 ,so (**) converts to
  x^2 + 5  + x - 2 ≤ 8  
 [Also, don't write garbages like adding -8 on both sides]
 or x^2 + x -5 <= 0.
  Since [-1 +/- sqrt(21)]/2 are the two roots of x^2 + x -5 = 0  
  x^2 + x -5 <= 0 implies  [-1 - sqrt(21)]/2 <= x <= (-1 + sqrt(21))/2
  (between these two roots).

  But x >=2, and (-1 + sqrt(21))/2 < (1 + sqt(25)/2  = 2, a contradiction. 
  This means there is nosulution in this case.

 Case (i) if x - 2 < 0, then |x-2| = -x+2 ,so (**) converts to
  x^2 + 5  - x + 2 &#8804; 8 or
  x^2 - x  -1 <= 0.
  
 two roots of x^2 - x -1 = 0 are  [1 +/- sqrt(5)]/2  
 Hence,   (-1 - sqrt(5))/2 <= x <= (-1 + sqrt(5))/2
 Since x -2 < 0, so x < 2 and we see that (-1 + sqrt(5))/2 <2 [OK}
 
 The solution set is [(-1 - sqrt(5))/2,(-1 - sqrt(5))/2 ]
 or (-1 - sqrt(5))/2 <= x <= (-1 + sqrt(5))/2


 Kenny