We will learn about quadratic inequalities based on the quadratic equation ax2+bx+c = 0, where a≠0, a, b, c are real numbers. The inequalities are of the form ax2+bx+c = 0

The solutions are real and different if the discriminant b2-4ac > 0

Suppose α (alpha) and β (beta) are the solutions (roots, zeros) of

ax2+bx+c = 0

Then, we can write

ax2+bx+c = a(x - α) (x - β)

α,β are real α≠β

Suppose α <β.

Now assume x is moving from the left to the right along the real number line. Consider how the sign of ax2+bx+c
varies in the process.

At the extreme left, x < α < β, then (x -α) < 0; that is, the value is negative and (x - β) < 0; that is, the value is
negative.

In this region, a (x - α ) (x - β) > 0 since a*-ve *-ve = a * +ve

At x = α

a(x -α )(x - β) = 0

When α < x <β

(x -α )> 0 x -α is +ve

(x -α )> 0 x - βis -ve

∴ a(x -α ) (x - β) < 0

a* (+ve) * (-ve) = a * -ve

In this region, a(x -α ) (x - β) < 0

At x = β

a(x -α ) (x - β) = 0

To the right, β for when x >β >α

(x -α ) > 0 x -α is +ve

(x - β) > 0 x - β is +ve

a (x -α ) (x - β) > 0

a* (+ve) * (+ve) = a * +ve

So a(x -α ) (x - β) > 0 in this region.

These results are summarized in the figure given below.

ax2+bx+c>0 ax2+bx+c<0 ax2+bx+c>0 α                           β

At the points x = α, x = β

ax2+bx+c=0

If a x2+bx+c=0 a>o has real and unequal roots, β then when

x < α ⇒     ax2+bx+c>0

x < α ⇒     ax2+bx+c>0

x = α ⇒     ax2+bx+c=0

α < x < β ⇒     ax2+bx+c<0

x = β ⇒     ax2+bx+c=0

x >β ⇒     ax2+bx+c>0

Now suppose that ax2+bx+c=0 a > 0 has real and equal roots; that is,

α = β = r (say) then when

x< r ⇒     ax2+bx+c>0

x=r ⇒     ax2+bx+c=0

and x>r ⇒     ax2+bx+c>0

This is because

ax2+bx+c=(x-r )2

A square is always positive.

If, however, ax2+bx+c=0 (a > 0) has no real roots then ax2+bx+c>0.

for all real x.

Consider the following examples.

Example 1

Solve x2-6x+8>0 and mark the values on the real number line.

Solution I: x2-6x+8>0

⇔     x2-6x > -8

here a = 1, b = -6, c = 8

By adding (-b/2a) to both sides of the inequality, we get That is, the absolute value of x-3 should be greater than 1.

∴ There are two possibilities

x-3>1----------------------- (1)

-(x-3)>1-------------------- (2)

Consider x-3>1

⇒     x>1+3

⇒     x>4

Consider -(x-3)>1

⇒     -x+3>1

⇒     -x>1-3

⇒     -x>-2

⇒     -(-x) < -(-2)

⇒     x<2.

The solution set of x2-6x+8>0 is {x/x<2} U {x/x>4}

No value of x lies between 2 and 4.

(check it out)

Solution II: Alternate method

x2-6x+8>0

Factorizing the left-hand side of the inequality

x2-4x-2x+8>0

⇒     x(x-4)-2(x-4)>0

⇒     x(x-2) (x-3)>0

⇔     (x-2)<0 and (x-4)<0

or (x-2)>0 and (x-4)>0

If x-2<0 then x<2

If x-4<0 then x<4

or

If x-2>0 then x>2

If x-4>0 then x>4.

This is possible only if x<2 or x>4. Example 2

Solve x2-6x+5<0 and mark it on the number line.

Solution:

x2-6x+5<0

Factorizing the left-hand side of the inequality

x2-5x-x+5 <0

⇒     x(x-5)-1(x-5)<0

⇒     (x-1) (x-5)<0

⇒     (x-1)<0 and (x-5)>0

or

(x-1)>0 and     ⇒ (x-5)<0

If x-1<0     ⇒ x<1

If x-5>0     ⇒ x>5

Or

If x-1>0     ⇒ x>1

If x-5<0     ⇒ x<5

Or

If x-1>0     ⇒ x>1

If x-5<0     ⇒ x<5 #### Solve the following inequalities and mark them on the real number lines

1. x2-x-2<0

2. 7+10x+3x2<0

3. 2x2+3x+1>0

4. 3x2+5x+2>0

5. x2+2x-3<0

6. 2x2-x-15>0

1. Solution:

To solve x2-x-2<0

Factorizing the left-hand side of the inequality we get

x2-2x+x-2<0

⇒     x(x-2)+1(x-2)<0

⇒     (x+1)(x-2)<0

⇒     (x+1)<0 and (x-2)<0

Or

(x+1)>0 and (x-2)<0

If x+1<0 then x<-1

If x-2>0 then x>2

Or

If x+1>0 then x>-1

If x-2<0 then x<2. 2. Solution:

To solve 7+10x+3x2 <0

Factorizing the left-hand side of the inequality

7+7x+3x+3x2<0

⇒     7(1+x)+3x(1+x)<0

⇒     (7+3x) (1+x)<0

⇒     7+3x<0 and 1+x>0

Or

7+3x>0 and 1+x<0.

If 7+3x<0 then 3x<-7

x< -7/3

If 1+x>0 then x>-1

Or

If 7+3x>0 then 3x>-7

x>-7/3

If 1+x<0 then x<-1

Comparing with 3. Solution:

To solve 2x2+3x+1>0

Factorizing the left-hand side of the inequality

2x2+ 2x+x+1>0

⇒     2x(x+1)+1(x+1)>0

⇒     (2x+1) (x+1)>0

⇒     2x+1>0 and x+1>0

Or

2x+1<0 and x+1<0

If 2x+1>0 then 2x>-1

x>-1/2

If x+1>0 then x>-1

Or

If 2x+1<0 then 2x<-1

x<-1/2

If x+1<0 then x<-1.

Comparing with 4. Solution:

To solve 3x2+5x+2>0

Factorizing the left-hand side of the inequality

3x2+3x+2x+2>0

⇒     3x(x+1)+2(x+1)>0

⇒     (3x+2) (x+1)>0

⇒     3x+2>0 and x+1>0

Or

3x+2<0 and x+1 <0

If 3x+2>0 then 3x>-2

x>-2/3

If x+1>0 then x>-1

Or

If 3x+2 then 3x<-2

x<-2/3

If x+1<0 then x<-1.

Comparing with 5. Solution:

To solve x2+2x-3<0.

Factorizing the left-hand side of the inequality

x2+2x-3<0

⇒     x2+3x-x-3<0

⇒     x(x+3)-1(x+3)<0

⇒     (x-1)(x+3)<0

⇒     (x-1)<0 and x+3>0

Or

(x-1)>0 and x+3<0

If x-1<0 then x<1

If x+3>0 then x>-3

Or

If x-1>0 then x>1

If x+3<0 then x<-3.

Comparing with 6. Solution:

To solve 2x2-x-15>0.

Factorizing the left-hand side of the inequality.

2x2-x-15>0.

⇒     2x2-6x+5x-15>0

⇒     2x(x-3)+5(x-3)>0

⇒     (2x+5)(x-3)>0

⇒     2x+5>0 and x-3>0

Or

2x+5<0 and x-3<0

If 2x+5>0 then 2x>-5

x>-5/2

If x-3>0 then x>3

Or

If 2x+5<0 then 2x<-5

x<-5/2

If x-3 <0 then x<3.

Comparing with 