Find the number of integral values of x that satisfy|x2 + 3x - 1| <...
|x2 + 3x - 1| < 2|x| + 5
We have to find out the boundary points for each modulus and adjust the + or - sign after removing the modulus.
For the left expression, x
2 + 3x - 1y = x, the roots are
√13 ≈ 3.6
So, the approx roots are -3.3 and +0.3.
For the right side of the expression, the boundary point is 0
Let us denote the same in the number line.
For R1,
|x2 + 3x - 1| < 2|x| + 5
x2 + 3x - 1 < 2x + 5
x2+ x - 6 < 0
(x-2) (x +3) < 0
x ∈ (-3,2)
Overlapping region is
For R2,
|x2 + 3x - 1| < 2|x| + 5
-x2 - 3x + 1 < 2x + 5
-x2 - 5x - 4 < 0
x2 + 5x + 4 > 0
(x + 1) (x + 4) > 0
x ∈ R −(−4,−1)
Overlapping region is
For R3,
|x2 + 3x - 1| < 2|x| + 5
-x2 - 3x + 1 < - 2x + 5
- x2 - x - 4 < 0
- x2 + x + 4 > 0
This is possible for all values of x. Therefore the overlapping region is
For R4,
|x2 + 3x - 1| < 2|x| + 5
-x2 - 3x + 1 < - 2x + 5
x2 + 5x - 6 < 0
(x + 6) (x -1) < 0
x ∈ (−6, 1)
Hence, the overall overlapping region is
Integral values are {-5, -4, -3, -2, -1, 0, 1}