# Quadratic Inequation and Steps to solve the inequalities in one variable for Class 11 ,CBSE Board, IITJEE maths and other exams

We have learned by Quadratic equation is previous chapter
so for a $\neq$ 0
ax2+bx+c > 0
or
ax2+bx+c < 0
or
ax2+bx+c $\geq$ 0
or
ax2+bx+c $\leq$ 0
are called Quadratic Equation in one Variable

## Steps to solve the inequalities in one variable.

1)  Obtain the linear inequation
2)  Pull all the terms having variable on one side and all the constant term on another side of the inequation
3) Simplify the equation in the form given above
ax> b
or
$ax \geq b$
or
ax< b
or $ax \leq b$
4) Divide the coefficent of the variable on the both side.If the coefficent is positive,direction of the inequality does not changes,but if it is negative, direction of the inequation changed
5) Put the result of this equation on number line and get the solution set in interval form
Example:
x- 2 > 2x+15
Solution
x-2 > 2x+15
-17  > x
Solution set
$(-\infty,-17)$

Some Problems to practice
1. 2x > 9
2) x + 5 > 111
3) 3x < 4,
4) 2(x + 3) < x+ 1

## Steps to solve the inequality of the form

(ax+b/cx+d)  > k or similar type
1)  take  k on the LHS
2) Simplify LHS to obtain the inequation in the form
(px+q/ex+f)  > 0
Make the coefficent positive if not
3)  Find out the end  points solving the equatoon px+q=0  and ex+f=0
4) Plot these numbers 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

Or there is more method to solves these

1)  take  k on the LHS
2) Simplify LHS to obtain the inequation in the form
(px+q/ex+f)  > 0
Make the coefficent positive if not
3)  For the equation to satisfy both the numerator and denominator must have the same sign
4) So taking both the part +, find out the variable x interval
5) So taking both the part -, find out the variable x interval
6) Write down the solution set in interval form
Lets take one  example to clarify the points
(x – 3)/(x + 5)> 0.
Solution
Method A
1) Lets find the end points of the equation
Here it is clearly
x=3 and x=-5
2) Now plots them on the Number line
3) Now lets start from left part of the most left number
i.e
case 1
x < -5 ,Let takes x=-6 (-6-3)/(-6+5) > 0
3 > 0
So it is good
Case 2
Now take x=-5
as x+5 becomes zero and we cannot have zero in denominator,it is not the solution
Case 3
Now x > -5 and x < 3, lets take x=1 (1-3)/(1+5) > 0
-1/6 > 0
Which is not true
Case 4
Now take x =3,then
0> 0 ,So this is also not true
case 5
x> 3 ,Lets x=4
(4-3)(4+5) > 0
1/9 > 0
So this is good
So the solution is
x < -5 or x > 3
or
$(-\infty,-5)\cup (3,\infty)$
Method B
1) the numerator and denominator must have the same sign. Therefore, either
1) x – 3 > 0 and x + 5 > 0,
or
2) x – 3 < 0 and x + 5 < 0. Now, 1) implies x > 3 and x > -5.
Which numbers are these that are both greater than 3 and greater than -5?
Clearly, any number greater than 3 will also be greater than -5. Therefore, 1) has the solution
x > 3.
Next, 2) implies
x < 3 and x < -5.
Which numbers are these that are both less than 3 and less than -5?
Clearly, any number less than -5 will also be less than 3. Therefore, 2) has the solution
x < -5.
The solution, therefore, is
x < -5 or x > 3