Modulus inequalities :
Modulus inequalities |x| < 1 =>
|x| > 1 => |a| = b
=> + a = b -1 < x < 1
x < -1 or x > 1
Quadratic inequalities :
Quadratic inequalities Quadratic inequality
(x - a)(x – b) < 0 =>
where a is smaller
than b
(x – a)(x – b) > 0 = > a < x < b,
x > b or x < a. < >
Solve |x – 3| < 7 :
Solve |x – 3| < 7 +(x – 3) < 7 (or) –(x – 3) < 7
x < 10 -4 10 –x + 3 < 7
-x < 4
x > -4 Answer is -4 < x < 10.
Solve |2x – 1| > 7 :
Solve |2x – 1| > 7 (or) +(2x – 1) > 7
2x > 7 + 1
2x > 8
x > 4 -(2x – 1) > 7
-2x + 1 > 7
-2x > 7 - 1
-2x > 6
x < -3 -3 4 Answer is
x < -3 (or)
x > 4
Solve 3x + 1 > |x – 5| :
Solve 3x + 1 > |x – 5| (or) 3x + 1 > - (x – 5)
3x + 1 > - x + 5
4x > 4
x > 1 3x + 1 > +(x – 5)
3x + 1 > +x – 5
2x > - 6
x > - 3 0 2 CHECK FOR x = 0
3x + 1 > |x – 5|
3(0) + 1 > |0 – 5|
1 > 5 (FALSE)
Since it’s wrong, therefore
Reject x > -3
Therefore answer is
x > 1
Solve |x – 3| > 2x + 1 :
Solve |x – 3| > 2x + 1 +(x – 3) > 2x + 1
+ x – 3 > 2x + 1
-x > 4
x < -4 (or) -(x – 3) > 2x + 1
- x + 3 > 2x + 1
-3x > - 2
x < -4 2
3 0 -5 CHECK FOR x = 0
|x – 3| > 2x + 1
|0 – 3| > 2(0) + 1
3 > 1 (TRUE)
Since it’s correct, therefore
x < 2
3
Solve |2x - 3| < |x + 1| :
Solve |2x - 3| < |x + 1| |a| = |b| => <
Solve |2x - 3| > |x + 1| :
Solve |2x - 3| > |x + 1| |a| = |b| => >