Linear Inequalities SS1 Further Mathematics Lesson Note

LINEAR INEQUALITIES IN ONE VARIABLE

Most of the rules for solving linear inequalities in one variable are similar to those for solving a linear equation in one variable except the rules on  multiplication and division  by negative number which reverses the sense of the inequality

EXAMPLE: Find the solution set of each of the following inequalities and represent them graphically 

(a) 2x – 3 < x + 7 (b) 3x  + 4 > 1 – 2x

Solution

2x – 3 < x + 7

Adding 3 to both sides

2x < x + 10

Subtracting x from both sides

X < 10

3x + 4 > 1 – 2x

Subtracting 4 from both sides

3x > – 3 – 2x

Adding 2x to both sides

5x >  -3

Dividing both sides by 5

x > -3/5

QUADRATIC INEQUALITIES IN ONE VARIABLE

To find the solution sets, of the quadratic inequalities of the form, ax2 + bx + c ≥ 0 or ax2 + bx + c ≤0. Note the following 

1) If a>0 and b>0    then          a.b>0  

  or  a<0 and b<0     then          a.b>0

2) If a<0 and b>0    then         a.b< 0

   Or a>0 and b<0    then         a.b< 0

Worked examples

1) Find the solution set of   x2 + x – 6 > 0 

Solution 

x2 + x – 6 > 0 

(x – 2)( x + 3} > 0

  x – 2> 0  or  x + 3<0

  x >2   or      x < -3

x – 2 < 0    or  x +3>0

x < 2    or   x > -3

-3 < x < 2 

-3 < x < 2 

2)  Show graphically the solution Set of the inequality x2 + 3x – 4 ≤ 0

Solution 

X2 + 3x – 4 ≤ 0

X2 + 3x – 4 = 0

. (x – 1)(x + 4) ≤ 0

x – 1 ≤ 0   or   x + 4 ≥ 0

.x ≤ 1 or x ≥ -4

X – 1 ≥ 0   or x + 4 ≤ 0

X ≥ 1 or x ≤ -4 

X ≤ 1 or x ≥ -4 

– 4 ≤ x ≤ 1

Quadratic Inequality Curve

We recall that the graph of  f(x) = ax² + bx + c  is a parabola if D ≥ 0, the parabola crosses the axis at two distinct points, this fact can be used to solve the inequality ax2 + bx + c ≥ 0 or ax2 + bx + c ≤ 0

Worked examples

1) Determine the solution set of the inequality x2 – x – 10 < 2

  X2 – x – 10 – 2 < 0

    X2 – x – 12 <  0 

(x + 3)(x – 4)  < 0

x + 3 < 0 or x – 4 > 0

x < -3 or x > 4

x + 3 > 0 or x – 4 < 0

x > -3  or  x < 4

Using Parabolic curves   

Coordination of points at which the curve cuts the axis (x + 3)(x – 4) = 0

   X = -3 , x = 4

Example:

Find the solution of the inequality x2 – 2x – 3 ≥ 0

Solution 

x2 – 2x – 3 ≥ 0 

(x + 1)(x – 3) ≥ 0

x + 1 ≥ 0 or x – 3 ≥ 0

x ≥ -1 or x ≥ 3

(x + 1) ≤ 0 or (x – 3) ≤ 0

x ≤ -1 or x ≤ 3

Solution set  -1  ≤  x  ≤  3

ABSOLUTE VALUES

If a number x is positive or negative the absolute value of x is denoted as │x│. The absolute value of a number is the magnitude of the number regardless of the sign.

Worked examples

1) │2x – 3│≥ 4

      2x – 3 ≥ 4

      2x ≥ 4 + 3

      2x ≥ 7

      .x ≥ 7/2

      .x ≥ 3½

OR

      – (2x – 3) ≥ 4

      -2 x + 3 ≥ 4

      – 2x ≥ 4 – 3

      -2x ≥ 1

         .x ≤ -½  

2. Find the solution set of the inequality │x – 2│<│x + 3│

    Solution 

   │x – 2│<│x + 3│ 

    (x – 2)2< (x + 3)2  ≡ x2 – 4x + 4 < x2 + 6x + 9

        – 4x – 6x < 9 – 4

        – 10x < 5

       x > – 5/10

       x > -½

ASSIGNMENT 

1) Find the range of values of x for which 7x – 12 ≥ x2

2) For what values of x is 2×2 – 11x + 12 positive? 

3) Find the values of x satisfying: |3x – 2| ≥ 3|x – 1|

4) Find the value of the constant k for which the equation 2×2 + (k + 3)x + 2k = 0 has equal roots.