# Complexities of Absolute Value (Modulus):

The most important thing in Absolute value (Modulus) is its definition. Without its definition, modulus will not exist. In other words, the definition of Modulus is sacrosanct. Absolute value or Modulus is denoted by || sign.

### Definition:

|x|= x if x ≥ 0.

And |x|= –x, if x < 0.

So, the definition of modulus says that if any number is positive, let’s say ‘x’, then ‘x’ will come out of the modulus as it is.

While if ‘x’ is negative, then ‘x’ will come out of the modulus with a negative sign and it will become –x.

For example, |5| = 5 because number inside modulus is 5 which is a positive number, so when it comes out of mod symbol, it will remain as it is.

Another example: |–1.5|= 1.5, because number inside mod symbol is –1.5 which is a negative number, so when it comes out of mod symbol, it will become –(–1.5) = 1.5.

So, we can conclude that any number coming out of modulus symbol will be positive.

E.g. 1: Solve: |x| = –5.

We just discussed the point that anything coming out of mod symbol will always be positive, but in this question, it is given to be equal to a negative number.

If we plug any value of x, when it comes out of mod symbol, it will remain positive only and it cannot be equal to –5. Even if we plug a negative value of ‘x’, when it comes out of mod symbol, it will always be a positive number and never equal to –5.

That’s why the answer to this question is no solution.

E.g. 2: Solve: |x – 3|< 0.

Again, rather than using any theory, we should use the logic that any value coming out of mod symbol will always be positive, but the question is asking for any value coming out of mod symbol to be negative which is not possible.

I hope the logic has been understood. And this logic will be used in many places when graphs also have to be drawn.

E.g. 3: Solve: |x| = 5.

There are multiple ways of solving this.

First method: The easiest way of solving this problem is to substitute values of ‘x’ in positive and negative terms to see which value of ‘x’ on substituting in |x| will be equal to 5.

Answer is pretty simple: ‘x’ can take only two values i.e., +5 OR – 5.

Second method: This method is the standard method of solving. We should learn this method as all the problems including the difficult ones can also be solved using this method.

Whenever there is a term including variable (like x & y) inside the mod symbol, we can solve the problem by taking two cases.

1st case: If x ≥ 0, then term inside the mod will come out of mod as it is.

So, we get x= 5. But after getting the answer, we need to verify it with main condition. Our answer should satisfy the main condition. Our main condition is x ≥ 0, and we got the answer as x =5. The answer lies within the main condition, so our answer is correct.

2nd case: If x < 0, then term inside the mod will come out of the mod with a negative sign.

So, we get – x = 5; so, answer we get is x = –5. Again, we will have to check the answer obtained with the main condition. Our main condition tells us that x should be negative and the answer which we are getting is also negative, so answer obtained is correct.

So, final answer is x = 5 OR x = –5.

E.g. 4: Solve: |x –2| = 4.

Again, we will solve it by taking two cases.

1st case: if x – 2 ≥ 0, [remember whatever is inside the mod symbol, condition is applicable on that term]. So, if the term in the mod symbol is positive, then it will come out of mod symbol as it is.

So, we get x – 2 = 6. On solving, we get x = 6. Again after finding out the answer, we need to find out that answer obtained should lie within the main condition. Answer obtained is x = 6, which lies in the main condition range x ≥ 2. So, it’s perfectly fine.

2nd case: if x – 2 < 0, then term inside the mod symbol will come out of with a negative sign.

So, we get – (x – 2) = 4; on solving we get x = –2. And the answer obtained x = –2 lies within the main condition x < 2. So, answer obtained is perfectly fine.

Final answer is x = 6 OR – 2.

E.g. 5: Solve: x2 – 3|x| – 4 = 0.

Again, we can solve this problem by taking two cases.

1st case: If x ≥ 0, then it will come out of mod symbol as it is.

So, the equation becomes x2 – 3x – 4 = 0. On solving, we get x = 4 OR – 1. Answer obtained has to satisfy the main condition, so x = –1 will be rejected as our main condition is ‘x’ has to be positive. So, only answer valid is x = 4.

2nd case: If x < 0, then it will come out of mod symbol with a negative sign.

So, the equation becomes x2 + 3x – 4 = 0. On solving we get x = –4 OR 1. So, the answer obtained x = 1 is not valid as it does not satisfy the main condition x < 0.

So, final answer from both cases is 4 OR – 4.

E.g. 6: Solve: |x –1| ≥ 5.

First Case: If (x – 1) ≥ 0, then it will come out of the mod symbol as it is.

So, the equation becomes (x – 1) ≥ 5 = x ≥ 6.

But, the answer which we obtained should lie within the main condition.

Main condition is: x ≥ 1 and answer obtained is x ≥ 6.

So, answer is x ≥ 6.

Second case: If (x – 1) < 0, then it will come out of the mod symbol with a negative sign.

So, the equation becomes –(x – 1) ≥ 5 = x ≤ –4. Again the answer obtained needs to be verified whether it satisfies the main condition or not. Main condition is: x < 1 and answer obtained is x ≤ –4.

So, the answer satisfying these two conditions will be x ≤ –4.

So, finally answer from both cases will yield the complete solution to our question and answer is x ≥ 6 OR x ≤ –4.

And to get the final answer, we need to merge the answers obtained from two cases. Answers obtained from the two cases are –1 ≤ x < 1.5, –3.5 < x < –1.

On merging these two answers, we get final solution as – 3.5 < x < 1.5.

All these problems can also be solved using “Distance Method” which is far efficient than this method. But, we should know this method as all the concepts of AND, OR are clarified and it also helps in solving term involving quadratic in modulus.

# Alternate Meaning of Absolute Value:

Absolute value or Modulus is always considered as the distance from origin.

For e.g., |5| = 5 can be seen on the number line as the distance of 5 units from the origin 0.

If we take another example like |x| = 5; we just need to determine the term inside the modulus symbol will become zero for which value of ‘x’. The term inside modulus is ‘x’ which will become zero at x = 0. So x = 0 is the origin and 5 given in the question is the distance from the origin x = 0.

If we are standing at point ‘0’ and want to travel a distance of ‘5’, we can either go in right-hand direction or in left-hand direction. If we go in right-hand direction, then after traveling a distance of 5 units we will reach x = 0 + 5 = 5. And if we in left-hand direction, then after traveling a distance of 5 units we will reach x = 0 – 5 = –5.

So, answer is x = +5 or –5.

We should recall that by using standard method also we got the same answer.

We will try to solve all those problems which we solved using standard method by this new intuitive method.

E.g. 8: Solve: |x –3| = 6.

Again we know how to solve this problem by standard method, let’s try out this new method. We just need to determine where the term inside the modulus will become zero.

The term inside modulus is (x – 3) and it will become zero at x = 3. So, if we are standing at point ‘3’ and want to travel a distance of 6 units, we can go in either left or right-hand direction.

If we go in right-hand direction, then we will reach x = 3 + 6 = 9.

And if we go in left-hand direction, we will reach x = 3 – 6 = –3.

So, answer is x = 9 or –3.

E.g. 9: Solve: |2x – 2| = 10.

Now, this is a slightly different problem than the previous one as the coefficient of x in this problem in not 1. We can still solve this problem using new approach. But, we need to make one alteration to suit our need.

We can rewrite the problem as: 2|x – 1| = 10 as |x – 1| = 5. Now it becomes the problem of the previous pattern. Now, the origin is 1 and the distance to be travelled is 5.

If origin is 1 and if we go 5 units in right-hand direction, we will reach x = 1 + 5 = 6.

And if we travel 5 units in left-hand direction, we will reach x = 1 – 5 = – 4.

So, answer is x = 6 or – 4.

Alternate way:

|2x – 2| = 10; we can assume 2x as y.

So, question becomes |y – 2|= 10. Origin is 2 and we have to travel distance of 10 units. If we go in right-hand direction, we will reach y = 2 + 10 = 12, and when we go in left-hand direction, we will reach y = 2 – 10 = – 8.

But, we needed to find solution for x. We assumed 2x as y and y = 12 or – 8.

So, 2x = 12 or 2x = – 8 which gives us the values of x as 6 or – 4. So, whenever the coefficient of ‘x’ is not 1, we need to make some adjustments.

E.g. 10: Solve: |3x + 6|= 10.

Again, we need to make the coefficient of ‘x’ one by extracting 3 common from modulus symbol.

So, we can rewrite 3|x + 2| = 10.

So, |x + 2| = 3.33. Now, the origin is – 2 and distance traveled has to be 3.33 units. If we move in right direction, we will reach x = –2 + 3.33 = 1.33. And if we move in left-hand direction, we will reach x =  –2 – 3.33 = –5.33.

So, answer is x = 1.33 or – 5.33.

E.g. 12: Solve: |x| ≤ 8.

If we know how to solve |x| = 8, we can solve this problem. Whenever an inequality problem comes in this format, we should first solve it first for equality, and then we can make necessary adjustments for inequality later on.

So, |x| = 8 can be solved orally as the origin is 0 and distance that needs to be traveled is 8. If we go in right-hand direction, we will reach x = 8, and if we go in left-hand direction, we will reach x =  –8.

So, answer for |x| = 8 is = 8 OR – 8.

Now, we need to solve for inequality. We need to revisit the logic we learnt in inequality that if there are two limits in inequality and inequality sign is (<), then the required answer will lie within the region. And if the inequality sign is (>), then answer lies outside the limits.

For e.g., (x – 2)*(x + 3) > 0, answer for this is x > 2 OR x < –3.

And if the question was (x – 2)*(x + 3) < 0, answer for this – 3 < x < 2.

So, same logic applies here also and we can solve all the inequality problems based on this concept.

So, coming back to our original question, if the answer for |x| = 8 is x = 8 OR – 8, then answer for |x| ≤ 8 will be –8 ≤ x ≤ 8. [There is < sign, so answer will lie within the limits].

E.g. 13: Solve: |x + 4| > 6.

Again an inequality sign is there in question, so we would solve it first for equality and then move onto inequality.

So, first of all, answer for |x + 4| = 6 can be calculated orally by assuming x = –4 as the origin and moving leftwards and rightwards 6 units to get x = –10 OR 2.

After finding this, the inequality sign given in question is (>), so answer will lie on the right-hand side of greater limit and on the left-hand side of lower limit.

So, final answer is x > 2 OR x < –10.

E.g. 14: Solve: |2x – 3| < 6.

We will solve it first for equal to, and then will make suitable adjustments for inequality sign.

So, |2x – 3| = 6 can be rewritten as 2|x – 1.5| = 6.

By canceling 2 on both sides, our question reduces to |x – 1.5| = 3. Now, the origin is 1.5, so by moving right-wards and left-wards 3 units, we will reach x = 4.5 OR – 1.5.

So, the answer for |2x – 3| < 6 will be – 1.5 < x < 4.5.

E.g. 15: Solve: 3 < |x + 2| < 7.

We can split it into two parts. If a range is given as 3 < x < 7, then it can be split into two parts as x > 3 AND x < 7.

Similarly, this question can also be broken into two parts as |x + 2| > 3 AND |x + 2| < 7.

We can solve the first part by equating it to equal sign. So, |x + 2| = 3 has origin at x = –2 and distance traveled has to be 3 units. So, x = 1 OR –5. The inequality sign in first part is greater than (“>”), so answer region will lie outside the limits. Finally answer for |x + 2| > 3 is x > 1 OR x < –5.

Second part can also be solved by equating it to equal sign. So, |x + 2| = 7 has origin at x = –2 and distance traveled has to be 7 units. So, x = 5 OR – 9. The inequality sign in this part is less than (“<”), so answer region will lie within the limits. So, answer for second part is –9 < x < 5.

Finally, complete answer for |x + 2| > 3 AND |x + 2| < 7 can be found out by finding out common solution since AND is present.

So, after finding common answer for [x > 1 OR x < –5] AND [–9 < x < 5], we get final solution as {–9 < x < –5 OR 1 < x < 5}.

All these questions are mixture of application of inequality and modulus. So, we need to have our fundas clear in both chapters to solve such problems.

E.g. 16: Solve: |x2 – 4x| < 2x.

The most efficient way of solving these type of problems is based on same logic as that of |x| < a. Answer for {|x| < a} should be solved orally as we have done these types of questions earlier. Answer for {|x| < a} is –a < x < a.

So, using the same funda, we can proceed for this question in same manner.

–2x < x2 – 4x < 2x can be again broken down into –2x < x2 – 4x AND x2 – 4x < 2x.

We will solve both parts separately and then find an intersection of solutions obtained from both parts as AND condition has to be satisfied.

First part: 0 < x2 – 4x + 2x can be solved as x2 – 2x > 0. Solution for this is x > 2 OR x < 0.

Second part: x2 – 4x < 2x can be rewritten as x2 – 6x < 0. Solution for this is 0 < x < 6.

Now, we need to find a common solution for these two answers to get final solution.

So, final answer is 2 < x < 6. {Apply AND for two solutions to get common solution}.

E.g. 17: Solve: |x2 + x – 7| > 5.

Again, this question is based on the logic of {|x| > a}. Answer for {|x| > a} is x > a OR x < –a.

The answer for all these results is based on same logic as if |x| = a, then solution will be x = a or – a. But, since ‘>’ sign is involved, required answer region will lie outside the solution region, thus arriving at the answer of x > a OR x <– a.

So, the solution set for question can be written directly as x2 + x – 7 > 5 OR x2 + x – 7 < –5.

Both the parts can be solved simultaneously as x2 + x – 7 – 5 > 0 OR x2 + x – 7 + 5 < 0.

On solving we get, (x + 4) (x – 3) > 0 OR (x + 2) ( x – 1) < 0.

So, answers for both the parts are [x > 3 OR x < –4] OR [–2 < x < 1]. Now, we need to apply OR to both solutions to get the final answer.

Final answer is [x > 3 OR x < –4] OR [–2 < x < 1].

1. a) x > 7 OR x < –11       b) No solution    c) x = – 2.5           d) x ≥ 3.66 OR x ≤ 0.33

2. 10                     3. 7                       4. 15                     5. All real values of ‘x’ except 0.

6. 2                       7. 8                       8. 7

This entry was posted in Algebra, Quantitative Aptitude and Mathematics and tagged , , , , . Bookmark the permalink.

### 16 Responses to Complexities of Absolute Value (Modulus):

1. prateekkanther says:

Done sir..!!

2. Amit Singh says:

done…..

1) a) x>7 or x=3.66 or x<=0.33

2)10

3)7

4) doubt

6)2

7)8

8)7

3. avinashgarg says:

done,,,, doubt on 4 & 7.

4. annkitsethi says:

Done

5. komalsaboo says:

1.a)x>7 OR x=3.66 OR x<=0.33
2)10
3)9
4)
5)All real values of ‘x’.
6)2
7)10
8)

6. raju.verma says:

1-a x>7 or x<-11
b- no salution
c- x=3.66 or X>=0.33
2- 10 values
3- 7 values
4-
5- all values
6- 2 values
7- 8 values

7. bhagyashree says:

All done …. Doubt in question 4 ,8

8. sumit_ag says:

1. a) x > 7 OR x < –11
b) No solution
c) x = – 2.5
d) x ≥ 3.66 OR x ≤ 0.33
2. 10
3. 7
5. All real values of ‘x’.
6. 2
7. 8
8. 7

9. pramita says:

1a)x>7 OR x<–11
1b)No solution
1c)x=–2.5
1d)x≥3.66 OR x≤0.33
2)10
3)7
4)15
5)All real values of ‘x’.
6)2
7)8
8)7

10. rohit.nema says:

A-1)a)x>7 or x3.66 or x<-.33
A-2) 10 values
A-3) 7 values
A-5) All real value of x
A-6) 2 values
A-7) 8 values
A-8) 7 values

1). a) x > 7 OR x < –11 b) No solution c) x = – 2.5 d) x ≥ 3.66 OR x ≤ 0.33
2)10
3)7
4)15
5)All real values of ‘x’.
6)2
7)8
8)7

12. chandan says:

1a)x>7 OR x<–11
1b)No solution
1c)x=–2.5
1d)x≥3.66 OR x≤0.33
2)10
3)7
4)15
5)All real values of ‘x’.
6)2
7)8
8)7

13. Snehal Chhajed says:

done

14. sushantsd says:

x>7 or x=3.66 or x<=0.33
10
7
15
all values of x
2
8
7

15. vi7vek says:

1a)x>7 OR x<–11
1b)No solution
1c)x=–2.5
1d)x≥3.66 OR x≤0.33
2)10
3)7
4)15
5)All real values of ‘x’, except 0
6)2
7)8
8)7

16. aniruddha says:

completed