- What are inequalities
- |
- Things which changes the direction of the inequality
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- Linear Inequation in One Variable
- |
- Linear Inequation in Two Variable
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- Quadratic Inequation
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- Steps to solve the inequalities in one variable
- |
- Steps to solve the inequality of the another form
- |
- Steps to solve Quadratic or polynomial inequalities
- |
- Absolute value equation
- |
- Absolute value inequation
- |
- Graphical Solution of Linear inequalities in Two Variable

ax^{2}+bx+c > 0

or

ax^{2}+bx+c < 0

or

ax^{2}+bx+c $\geq$ 0

or

ax^{2}+bx+c $\leq$ 0

1) Obtain the Quadratic equation inequation

2) Pull all the terms having on one side and Simplify the equation in the form given above

3) Simplify the equation in the form given above

4) Find the roots of the quadratic equation using any of the method and write in this form

(x-a)(x-b)

5) Plot these roots on the number line .This divide the number into three segment

5) Start from LHS side of the number and Substitute some value in the equation in all the three segments to find out which segments satisfy the equation

6) Write down the solution set in interval form

**Example**

1) Simpify the inequality which means factorizing the equation in case of quadratic equalities

example

x2-5x+6 > 0

Which can be simpified as

(x-2)(x-3) > 0

2) Now plot those points on Number line clearly

3) Now start from left of most left point on the Number line and look out the if inequalities looks good or not. Check for greater ,less than and equalities at all the end points

So in above case of

x2-5x+6 > 0

We have two ends points 2 ,3

Case 1

So for x < 2 ,Let take x=1,then (1-2)(1-3) > 0

2 > 0

So it is good

So This inequalties is good for x < 2

Case 2

Now for x =2,it makes it zero,so not true. Now takes the case of x > 2 but less 3.Lets takes 2.5

(2.5-2)(2.5-3) > 0

-.25 > 0

Which is not true so this solution is not good

Case 3

Now lets take the right most part i.e x > 3

Lets take x=4

(4-2)(4-3) > 0

2> 0

So it is good.

Now the solution can either be represented on number line or we can say like this

$(-\infty,2)\cup (3,\infty)$

1) Solving them is same as steps given above to solve the linear inequation and quadratic equation

2) Solve both the inequation in the pair

3) Look for the intersection of the solution and give the solution in interval set

1) We cannot have zero in denominator

2) We should be checking for equalities at all the end points

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