so for a $\neq$ 0

$ax^2+bx+c > 0$

or

$ax^2+bx+c < 0$

or

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

or

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

are called Quadratic Inequation in one Variable

or

$ax^2+bx+c < 0$

or

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

or

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

- Obtain the Quadratic inequation
- Pull all the terms having on one side and Simplify the equation in the form given above
- Find the roots(0 points) of the Quadratic equation using any of the method and write in this form

$(x-a)(x-b) $ - Plot these roots on the number line .This divide the number into three segment
- 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

- Write down the solution set in interval form

$x^2-5x+6 > 0 $

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

Which can be simpified as

$x^2-5x+6 > 0 $

$x^2 -2x-3x+6 > 0$

$(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

$x^2-5x+6 > 0$

We have two ends points 2 ,3

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

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

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

$-.25 > 0$

Which is not true so this solution is not good

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

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

- $x^2 -7x +10 > 0 $
- $x^2 -2x -15 < 0 $
- $x^2 +2x +1 > 0 $
- $x^{2} - 13 x + 36 > 0$
- $x^{2} - 8 x + 7 < 0$
- $x^{2} - 12 x + 20 \geq 0$
- $x^{2} - 11 x + 28 \leq 0$
- $x^{2} - 6 x + 8 \geq 0$
- $x^{2} - 8 x + 12 \leq 0$
- $x^{2} - 16 x + 64 \geq 0$
- $x^{2} - 5 x + 6 \leq 0$
- $x^{2} - 18 x + 80 > 0$
- $x^{2} - 4 x + 3 < 0$$

so for a $\neq$ 0

$ax^3+bx^2+cx+d > 0$

or

$ax^3+bx^2+cx+d< 0$

or

$ax^3+bx^2+cx+d \geq 0 $

or

$ ax^3+bx^2+cx+d \leq 0$

are called Cubic Inequation in one Variable

or

$ax^3+bx^2+cx+d< 0$

or

$ax^3+bx^2+cx+d \geq 0 $

or

$ ax^3+bx^2+cx+d \leq 0$

- Obtain the cubic inequation
- Pull all the terms having on one side and Simplify the equation in the form given above
- Find the roots(0 points) of the cubic equation using any of the method and write in this form

$(x-a)(x-b)(x-c) $ - Plot these roots on the number line .This divide the number into Four segment
- 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

- Write down the solution set in interval form

- Solving them is same as steps given above to solve the linear inequation and quadratic inequation
- Solve both the inequation in the pair
- Look for the intersection of the solution and give the solution in interval set

- We cannot have zero in denominator

- We should be checking for equalities at all the end points

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

**NCERT Solutions**