SOLUTION: I have been working on this problem forever and can not come up with the answer in the back of my book.
Please help me solve the following equation :
│x^2 + 5│
Algebra.Com
Question 7071: I have been working on this problem forever and can not come up with the answer in the back of my book.
Please help me solve the following equation :
│x^2 + 5│ + │x - 2│ ≤ 8
This is what I did .
│x^2 + 5 + x - 2│≤ 8
-8 ≤ x^2 + 5 + x – 2 ≤ 8
-8 ≤ x^2 + x + 3 ≤ 8
then subtract -3 from all sides and get this
-11 ≤ x^2 + x ≤ 5, Then completing square for middle getting this
-11 + 1/4 ≤ x^2 + x + 1/4 ≤ 5 + 1/4
-43/4 ≤ x^2 + x + 1/4 ≤ 21/4
-43/4 ≤ (x + 1/2)^2 ≤ 21/4
Then don’t go any further because I now that I can not get the answer that is in the book.
Please tell me where I am going wrong.
TIA,
del
Answer by khwang(438) (Show Source): You can put this solution on YOUR website!
│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 ≤ 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
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