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Quadratic And Cubic inequalities





Quadratic Inequation

We have learned by Quadratic equation is previous chapter
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 Equation in one Variable

Steps to solve Quadratic or polynomial inequalities

$ax^2+bx+c > 0$
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. Find the roots(0 points) of the quadratic equation using any of the method and write in this form
    $(x-a)(x-b) $
  4. 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

Solved Example


Question 1
$x^2-5x+6 > 0 $
Solution
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
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 $2< x < 3$. Lets takes 2.5
$(2.5-2)(2.5-3) > 0$
$-.25 > 0$
Which is not true so this solution is not good
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
$(-\infty,2)\cup (3,\infty)$

Practice Questions
  • $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$$

Cubic Inequation

We have learned by Quadratic equation is previous chapter
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 Quadratic Equation in one Variable

Steps to solve Cubic inequalities

$ax^3+bx^2+cx+d > 0$
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$
  1. Obtain the cubic equation inequation
  2. Pull all the terms having on one side and Simplify the equation in the form given above
  3. Find the roots(0 points) of the vunic equation using any of the method and write in this form
    $(x-a)(x-b)(x-c) $
  4. Plot these roots on the number line .This divide the number into Four 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

Practice Questions with hints

  • $x^{3} - 17 x^{2} + 86 x - 112 > 0$
  • $x^{3} - 14 x^{2} + 64 x - 96 < 0$
  • $x^{3} - 12 x^{2} + 29 x - 18 > 0$
  • $x^{3} - 13 x^{2} + 32 x - 20 > 0$
  • $x^{3} - 17 x^{2} + 76 x - 60 < 0$
  • $x^{3} - 27 x^{2} + 242 x - 720 > 0$


Steps to solve the pair of Linear inequation or quadratic equation

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

Some Important points to note for inequalities

  • We cannot have zero in denominator
  • We should be checking for equalities at all the end points

Related Topics


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