Demystifying Inequalities:

What is basic difference between solving an Inequality and Equality problem?

Answer is pretty simple. While solving an equality problem, we get a fixed value/s of the variable but in case of an inequality, we get a range of values.

For e.g., On solving 2x – 5 = x – 3, we get value of the variable x as 2.

But the same problem with inequality will give us range of values. 2x – 5 > x – 3; On solving this, we get x > 2. The range signifies that ‘x’ can take all the values which are greater than 2 and uptill + infinity.

Let us see pictorial understanding of ‘>’ and ‘<’.

So, throughout the inequality chapter we will use the notation “right of” or “left of” particular value instead of greater or less than that particular value. This concept helps us in visualizing the number line and gives us our required region.

AND/ OR Funda:

We use funda of AND/OR quite extensively without paying attention to its details.

For e.g., if there are two sets given, let’s say.

{Classical Batsmen} = [Dravid, Kallis, Sachin, Mahela, Ponting, Clarke].

{Hard–hitting batsmen} = [Hayden, Sehwag, Sachin, Kallis, Ponting].

If now, the question arises which batsmen will comprise the set who are Classical AND hard–hitting batsmen. I am sure all of us will answer this. We will find out such names in two sets which are common to both sets. [Kallis, Sachin, Ponting].

Similarly, if the question asks which batsmen will comprise the set who are classical OR hard–hitting?

Answer will be the set of all the batsmen who are uniquely present in both the sets. [Hayden, Sehwag, Sachin, Kallis, Ponting, Mahela, Dravid, Clarke].

One more pointer in AND/OR case is that OR set will always be greater than AND set.

OR (Union): In case of OR, either of the conditions has to be satisfied. So, even if one condition is satisfied, that element will be included in OR. In the above example, if any of the batsmen satisfy condition of hard–hitting or classical, they will be included in the set.

AND (Intersection): In case of AND, both the conditions have to be satisfied simultaneously. If only one of the conditions is being satisfied, then that case will not be considered. In the above example, if we are looking for set of classical batsmen AND hard–hitting batsmen, then both the conditions have to be met at same time.

Application of AND/OR in Inequality:

AND/OR is used quite frequently in inequality problems, so we will revisit this funda again. But this time we will play with inequality signs.

E.g. 1: If x > – 2 AND x < 3.

E.g. 2: If x > – 2 AND x > 3.

E.g. 3: If x > –2 OR x > 3.

E.g. 4: If x < –2 AND x > 3.

E.g. 5: If x < –2 OR x > 3.

These many examples should be enough to understand the logic of AND/OR. Best way to remove any doubts related with this logic is again OR set will be greater than AND set.

SOME RULES OF INEQUALITY:

A) Addition or subtraction: Addition or subtraction in case of inequality is exactly same as equality problems. There is no change. We can add/subtract without any worry.

For e.g., if a < b, if we add k (k can be a positive number or a negative number) on both sides, inequality sign will remain same.

I want to say that a + k < b + k.

Take values and verify it for yourself.

We know 5 < 8, if we add –4 on both sides we get (5 – 4 < 8 – 4) = 1 < 4.

Inequality sign still holds even if we add a negative number on both sides.

B) Multiplication or Division: We have to tread this path carefully as there are some restrictions in this area. First of all, when we are multiplying or dividing by a number in inequality problems, we should know the polarity (sign) of that particular number. If we don’t know the sign, we can’t multiply or divide blindly. If we are multiplying or dividing by a positive number, then inequality sign will hold true. But if we are multiplying or dividing by a negative number, then inequality sign will reverse. We have to be utmost careful here as this is an error–prone area.

For e.g., (i) if a < b and we are multiplying by a number ‘k’ on both the sides and ‘k’ being a positive number, inequality sign will not change.

It means we will get a × k < b × k.

We can verify it by taking values, If 5 < 7 then multiply by 2 on both sides. ( 2 is a positive number).

Inequality sign will hold true as we get 10 < 14.

(ii) if a < b and we are multiplying by a number ‘k’ on both the sides with ‘k’ being a negative number, inequality sign will reverse.

It means we will get a × k > b × k.

If we take values like 3 < 5 and multiply by –2 on both sides, inequality sign will reverse as we will get –6 > –10 . (–2 is a negative number).

Same logic holds true for division also.

C) Squaring and Cubing: We do not have issues with cubing. Cubing both sides with any inequality sign is tension–free. So, we don’t need to worry when we need to cube both sides as inequality sign remains the same.

For e.g. if a > b, then a³ > b³ (irrespective of the sign of a & b).

In case of squaring, we need to be wary of signs. We will take three cases to check all the possibilities.

(i) When both the numbers are positive; on squaring inequality sign remains same.

For e.g., If a > b, then a² > b² (if a & b are +ve numbers).

If 4 > 3, then on squaring sign will hold true as we will get 16 > 9.

(ii) When both numbers are negative; on squaring inequality sign will reverse.

If a > b, then a² < b² (if a & b are negative numbers).

For e.g., if –3 > – 5, then on squaring inequality sign will reverse as we will get 9 < 25.

(iii) When one number is positive and other one in negative, nothing can be said of sign. It will depend on the magnitude.

For e.g., if 5 > – 3, then on squaring we get 25 > 9. In this case the inequality sign holds true as the magnitude of 5 is greater than 3.

But if 5 > – 6, then we will get 25 < 36 on squaring as the magnitude of 6 is greater than 5.

So, when one of the numbers is +ve and another one is –ve, we will have to look out for magnitude and accordingly proceed.

There are many other rules in inequality which we will come to know when we solve different problems. We will learn those theories when we encounter those problems.

Applications of Inequalities:

E.g. 6: Solve: 2x + 5 > 2(x + 1).

This is a simple problem based on the rules of addition/subtraction.

We can rewrite this problem as 2x + 5 – 2x > 2.

On solving we get, 5 > 2.

We are getting an answer independent of ‘x’. Whenever this happens, we need to check the condition whether its true or false.

Condition we are getting is 5 > 2. Is it true?

Yes, it is true. So, answer should be x can take all real values.

We could have also approached this problem in a different way.

Question is 2x + 5 > 2x + 2. Now, 2x is on both sides and 2x added to 5 will always be greater than 2x added to 2 for any real value of x. That’s why the answer is all real values of x.

Now, we will look at how to tackle higher–degree inequalities like quadratic, cubic and so on.

E.g. 7: Solve: x² – x – 6 > 0.

There are two ways of solving this problem. First way is the graphical method which we will see later and the second method is the polarity method which we will learn now. All higher–degree inequalities can be solved using POLARITY METHOD. Let’s see how to do

POLARITY METHOD:

x² – x – 6 > 0 can factorised by splitting the middle terms as (x – 3) × (x + 2) > 0.

Two numbers i.e. (x – 3) and ( x + 2) are being multiplied to result in a positive number. (Start reading > 0 as positive). Think what are the cases possible for two numbers being multiplied and resulting into a positive number.

First case: Both numbers are positive i.e. both(x – 3) and (x + 2) are positive.

Second Case: Both (x – 3) and (x + 2) are negative.

Now, next step is to plot points on the number line where (x – 3) and (x + 2) will become zero. (x – 3) will become zero at x = 3 and (x + 2) will become zero at x = –2.

So, entire number line is divided into three–regions. Those regions are x > 3, –2 < x < 3 and x < –2.

If we take any value in the first region which is x > 3, then both the brackets (x – 3) and (x + 2) will be positive and two positive numbers multiplied will be a positive number.

Take a value for x in the region x > 3. If x = 5, then (x – 3) (x + 2) > 0 is true as both the brackets will be positive and when multiplied will give us the required result. So, we can say that x > 3 is one of the regions which satisfy our requirement.

If we take the middle region which is –2 < x < 3, then one of the brackets will be positive and other will be negative; and multiplication of positive number with a negative number results in a negative number. Take a value for x in the region –2 < x < 3. If x = 0, then value of (x – 3) becomes –3 and value of (x + 2) becomes 2. And when we multiply these two, we get a negative number. So, this middle region does not suit our requirement.

If we take the leftmost–region which is x < –2 , then both the brackets will be negative and two negative numbers multiplied always give a positive resultant.

Take a value in the region x < –2. If x = –4, then value of (x – 3) becomes –7 and value of (x + 2) becomes –2; when these two multiplied, we get a positive number which fulfils our need.

This pattern will always follow. That is, in the right–most region, we got positive result; in the middle–region, we got negative result; and in the left–most region we got positive result. Observe that the polarity sign changes alternately if we start from right–most region to left–most region. We will use this funda through–out the chapter. But, what is the logic behind this funda?

The polarity funda is based on this underlying logic of left–right of a point. This logic is very useful and will be used in subsequent chapters also. And in all the inequality problems, we will always start from the right–most region because we will always get positive region. And after that we don’t need to do anything as sign changes alternately.

So, you should have got the idea that for a quadratic inequality, if the question is asking for > 0, then answer will always lie beyond the boundaries and if the question is asking < 0, then our required region will be with–in the boundaries. See the picture above to get the logic.

And if we have understood this logic, we can solve these questions orally by just finding out the limiting points (points where that particular bracket will be zero).

E.g. 8: Solve: x² – x – 6 < 0.

Same question as the previous one, only difference is that we need to find range of values for x for which the quadratic expression given in the question will be negative.

Again, we will factorize the expression, we get (x – 3) ( x + 2) < 0.

We will find out the limiting–points or boundaries where the two brackets will become zeros.

The points are same as the previous one and since the question is asking for the negative result, our required region will lie between the boundaries.

So, answer is – 2 < x < 3.

E.g. 9: Solve x² – x – 6 ≥ 0.

This question is similar to x² – x – 6 > 0, only difference being that in previous question the value of the quadratic expression had to be greater than zero, but in this question the value of quadratic expression has to be either equal to zero or greater than zero.

So, the boundaries will still remain the same, means the boundaries are still 3 and –2. But, in this question the points 3 and –2 will also be included in the answer. Because when x takes the value of –2, the value of the quadratic expression (x – 3) ( x + 2) ≥ 0, will be zero and when x takes the value of 3, the value of the expression will be zero.

So, answer will be x ≥ 3 or x ≤ –2.

When there is an equal to sign with an inequality sign (i.e. ≥ or ≤), we don’t need to do anything different in finding out the required region. Just plot the points and if equal to sign is there, then those points will also be included in our answer.

E.g. 10: Solve (x + 5) (x + 7) ≤ 0.

First of all, we will find out the boundaries or points where the expression given becomes zero. Those points are –5 and –7 respectively.

Since, we are looking for the value of the expression to be negative; our answer will be within the region. Since, in the question equal to sign is also included the boundaries (i.e. –5 & –7 will also be included).

So, answer is –7 ≤ x ≤ –5.

So, all the expressions including two brackets or quadratic expression can be solved orally like we have learnt just now. Let us see a variation in this type.

E.g. 11: Solve –x² – x +12 > 0.

First of all, this problem is not in the form of the standard way. Whenever, the coefficient of x² is negative, first thing that we should do is to make the coefficient positive by multiplying the entire expression by –1. And we have learnt that when we are multiplying by a negative number, inequality sign reverses.

So, after multiplying, we get x² + x – 12 < 0, now this problem is in standard format. We can go ahead and solve this question with the logic learnt.

On factorizing, we get(x + 4) ( x – 3) < 0. So, the boundaries are (–4) and (3) respectively and whenever the quadratic expression is negative (less than zero), required answer lies within the region.

So, answer is [–4 < x < 3].

So, don’t forget to convert the coefficient of x² if its negative.

E.g. 12: Solve (x)(x – 2) ( x + 3) > 0.

This is a cubic inequality which can be solved using the polarity logic. Just plot the points on the number line where each of the brackets becomes zero. After that, we will start from the right–most region since we always get a positive result in that region. After that, we know that the sign changes alternately in the adjacent regions.

So, answer is x > 2 OR –3 < x < 0.

E.g. 13: Solve (2x + 3) ( x – 5) (x + 4) ≤ 0.

Again a cubic inequality, so we should find out the points where each of the brackets will become zero. First bracket will be zero when x = –3/2 or –1.5. Second bracket will become zero when x = 5 and third bracket will become zero at x = –4. Plot these points on the number line.

So, answer is –1.5 ≤ x ≤ 5 OR x ≤ –4. Notice, equal to sign is also included in the answer.

E.g. 14: For how many integral values of x, is the expression x(x – 5) (2x – 6) (x + 3) ≤ 0?

Most of the students get confused in these types of questions despite being based on the same logic. Only difference in this question compared to those which we have done is that in this question range is not being asked, but number of integral values of ‘x’ is being asked. If we could find the range for which the expression is less than or equal to zero, we can also count the number of integral values of ‘x’ for which this expression will be true.

Let us find out the range first.

The boundaries are 0, 5, 3 and –3 respectively. Plot them on number line and start from the right–most region because we always get positive there.

If range was being asked, answer is quite visible which is [3 ≤ x ≤ 5 OR –3 ≤ x ≤ 0].

But, since the number of integral values is being asked, we will count integers in the range found above. So, answer is 7 integral values. [Be careful, the boundaries will also be included in the answer as the value of the expression can be zero also].

Now, in this problem expression is being divided. How to solve this?

If the problem was (x – 2)(x – 5) > 0, we would have solved it orally and got the answer as x > 5 OR x < 2.

For, (x –2) (x – 5) > 0, how many scenarios are possible to multiply two numbers to get a positive result.

There are only two cases possible: a) Both numbers are negative or b) Both numbers are positive.

possible to divide two numbers to get a positive result. Again same two cases two possible; either both numbers are positive or negative.

That’s why the answer for both expressions is same.

Remember denominator cannot be zero at any cost. So, we will have to exclude x = 5 from our answer. But, in case of (x – 2) (x – 5) ≤ 0, there is no term in denominator, so we don’t have any restriction here.

So, whenever there is an equality sign with an inequality for an expression involving numerator & denominator, we will should be careful to exclude the value of ‘x’ for which denominator becomes zero.

Since, the question is asking for range of values for which the expression will be greater than or equal to zero, we can write the answer as [x > 2 OR –4 ≤ x ≤ –1].

Remember, denominator becomes zero at x =2, and since the question is asking for values for which the expression becomes greater than or equal to zero, we will have to exclude 2 from our answer.

Inequalities involving Perfect squares:

E.g. 18: Solve x² > 0.

We need to find the range of values of ‘x’ for which x² will be positive. We know x² is a perfect square and irrespective of whichever value of x, x² will always be positive. Even if we substitute the value of x as negative number, x² will be positive.

So, answer should be ‘x’ can take all real numbers except x = 0. ‘x’ cannot take value 0 because if we substitute x = 0 in x², value of x² will become 0 and the question is asking for greater than 0.

E.g. 19: Solve (x –2)² ≥ 0.

Again (x – 2)² is a perfect square and it will be positive for all real values of ‘x’. And in the question, we need to find values of ‘x’ for which the expression (x – 2)² will be greater than or equal to zero. So, x can take value of 2 also as in question equality sign is included.

So, the answer is all real values of ‘x‘.

E.g. 20: Solve (x + 7)² < 0

Answer should be no solution as (x + 7)² is a perfect square and it cannot be negative for any real values of x.

E.g. 21: Solve ( x + 5)² (x – 5) ≥ 0.

Now, this problem consists of two terms in which one of the terms is a perfect square. We can solve all such problems by ignoring the term involving perfect square as the perfect squares are always positive. So, answer is not dependent on them. Still if you are not clear with logic, we will go back to original logic of polarity.

Logic: When two terms are being multiplied to result into positive, only two cases are possible.

First case: Both the terms are negative. This case is not valid since the term (x + 5)² cannot be negative as it is a perfect square.

Second case: Both the terms are positive. This case is valid since (x + 5)² is positive. And since (x + 5)² is positive, second term also has to be positive.

So, our question reduces to (x – 5) ≥ 0. [That’s why i stated initially that we can ignore the perfect square as it is always positive, so answer is dependent on the second term only].

So, on solving answer is x ≥ 5, but we need to check that the perfect square which we ignored becomes zero at what point. And the question asks for expression to be greater than or equal to zero, so we will have to include that point in answer.

(x + 5)² becomes zero when x = –5 and we got our answer as x ≥ 5. X = –5 does not lie in the second region, so we will have to write it separately.

So, our final answer is x ≥ 5 and x = – 5.

E.g. 22: Solve (x –3)² ( x – 5) < 0.

Again the first term is a positive square, so we can ignore it initially as it is always positive.

So, question reduces to ( x – 5) < 0.

And answer is x < 5, again we need to check for the perfect square which we ignored where it becomes zero. The term ( x – 3)² becomes zero at x = 3 and since the question is asking for all values of ‘x’ for which the value of expression is less than zero and not equal to zero.

So, if x = 3 comes in our answer region, we will have to exclude it. On checking, we can see that x = 3 lies in the region x < 5, so we will have to exclude it.

So, final answer is x < 5 and x ≠ 3.

Again, the denominator is a perfect square, so we can ignore it and work on the numerator.

Our question reduces to (x – 2) ≥ 0. So, the answer is x ≥ 2. But, again we need to check where the denominator becomes zero. The denominator ( x – 3)² becomes zero x = 3. And the question is asking for values of ‘x’ for which the expression is greater than or equal to zero. But, we have to also take in mind that the (x – 3) ² is in denominator and it cannot be equal to zero as if the denominator becomes zero, the entire expression will become undefined.

So, if x = 3 comes in answer region x ≥ 2, we will have to exclude it. And x = 3 lies in the answer region. So, our final answer is x ≥ 2 and x ≠ 3.

So all the questions involving perfect squares can be solved pretty easily, but we just need to be careful in writing our final answer depending on whether the expression is asking greater than or equal to or less than zero.

Now, we need to check for the perfect square which we ignored. The term ( x – 2)² becomes zero at x = 2 and since the question is asking foe values of ‘x’ for which the expression is greater than or equal to zero, we will have to include x = 2 in our answer as perfect square is in numerator.

But, x = 2 does not lie in the answer region x > 3, so we will have to write it separately.

Final answer is x > 3 and x = 2.

E.g. 25: Solve (x – 5)³ (x + 3) ≥ 0.

Now, cube comes in picture, but don’t worry we can handle it by manipulating the question.

We can rewrite the question as (x –5 )² (x – 5) ( x + 3) ≥ 0.

Now, we know how to handle inequality with perfect square, we can ignore the perfect square and solve the remaining inequality.

Again, our question reduces to (x – 5) ( x + 3) ≥ 0.

We know how to find the answer of such inequalities, answer always lies outside the boundaries and the boundaries are + 5 and – 3.

So, answer is x ≥ 5 OR x ≤ – 3.

E.g. 26: Solve x² + x + 1 > 0.

There are two ways of solving this problem.

First Method: For any real value of x, x² will be always positive. And even if x is negative, x² will always positive and when x is added to x², x² + x will be still a positive number. And when we add 1 to x² + x, the resultant x² + x + 1 will always be positive.

So, we reached the conclusion that x² + x + 1 will always be positive for any real value of x and question is also asking for positive value. So, answer is all real values of x.

Second method: Whenever we encounter a quadratic expression, we try to split the middle term to find the roots. But, in this case we cannot split the middle term. So, here should immediately find out the determinant of the expression.

D = b² – 4ac= 1 – 4 = – 3.

If D is negative and coefficient of x² is positive, the graph of the expression y = x² + x + 1 will always be above the x–axis. It would not cut or touch the x–axis because if it touches or cuts the X–axis, then the roots become real which is a contradiction to the result obtained.

So, whenever, the quadratic expression cannot be factorized, we need to find the determinant immediately. If the determinant is negative and the coefficient of x² is positive, value of that particular quadratic expression is always positive.

E.g. 27: Solve (x² + 2) (x² –10x – 24) < 0.

Again the first term is a quadratic expression which cannot be factorized and if we find out its determinant, it is negative. So, x² + 2 is always positive as the roots are imaginary and the graph will always lie above the X–axis. So, we can ignore it completely as it would not have any effect on inequality.

So, our question reduces to x² – 10 x – 24 < 0 which can be further split into (x – 12) ( x + 2) < 0.

We know how to solve such inequalities orally and answer is –2 < x < 12.

When there is some other number on right hand side of inequality apart from zero:

E.g. 28: Solve ( x – 2)² >49.

Best way to solve such problems is to bring whatever is on the right–hand side to left–hand side because that is our standard format. If we do not do that, we will have to two cases to solve it which will consume slightly more time.

So, we can rewrite the question as (x – 2)² – 49 > 0.

Or (x – 2)² – 7² > 0; ( x – 2 + 7) ( x – 2 – 7) > 0.

We get (x + 5) ( x – 9) > 0.

So, the answer is x > 9 OR x < – 5.

Finding out minimum & maximum values for a given range:

E.g. 31: If 4 ≤ x ≤ 7, then find the maximum & minimum values of x².

This is a straight–forward problem whose answer will be 16 ≤ x² ≤ 49.

E.g. 32: If –5 ≤ x ≤ –2, then find the maximum & minimum values of x².

This is also a easy problem whose answer will be 4 ≤ x² ≤ 25.

E.g. 33: If –5 ≤ x ≤ 2, then find the maximum & minimum values of x².

Most of students will give answer of this question as 4 ≤ x² ≤ 25, which is wrong. This is one of the error–prone areas.

Range of x is from –5 to 2, it means that x can take all values from –5 to 2. x can take –1, 0 also. And if x = –1 and if we square it, we get x² = 1.

Similarly if we take x = 0, then x² = 0. These two values of x² i.e. 1, 0 should also be part of the range of x² which we got earlier 4 ≤ x² ≤ 25. But, 0 and 1 are not present in this range. That’s why the range obtained above is wrong.

Correct answer is 0 ≤ x² ≤ 25.

Whenever range of x is from negative to positive, lower limit of x² will always be 0 and upper limit of x² will depend on magnitude of upper & lower limit of x. For example, in this case the magnitude of –5 was greater than that of 2.

So, we need to be careful here. Whenever x lies between –1 & 0, range of x² always lies between 0 and 1.

E.g. 34: If –5 ≤ x ≤ 7, then find the maximum and minimum values of x².

Again the range of x is from negative to positive, the lower limit of x² will be 0 and upper limit of x² will depend on the magnitude of upper and lower limits of ‘x’. Magnitude of 7 is greater than that of –5, that’s why the upper limit of x² would be 49.

Answer is 0 ≤ x² ≤ 49.

E.g. 35: If 25 ≤ x² ≤ 64, find the range of x.

Most of us will give answer as 5 ≤ x ≤ 8. There is nothing wrong with this answer except that it is partial answer. Most of us don’t think that the negative part will also be included in answer.

Final answer is 5 ≤ x ≤ 8 OR –8 ≤ x ≤ – 5.

So, equation becomes y² + 10y – 24 ≤ 0.

We know how to solve such inequalities and answer is –12 ≤ y ≤ 2.

But, the question is asking for the range of x and not that of y. So, substitute the value of y in the inequality.

We get –12 ≤ √x ≤ 2.

Now to find the upper and lower limits of x, we will have to square the limits of √x, but the range of √x is passing from negative to positive, so the lower limit of x will be zero and upper limit of x will depend on the magnitude of lower & upper limit of √x.

So, finally answer is 0 ≤ x ≤ 144.