LINEAR INEQUALITIES

LINEAR INEQUALITIES IN ONE VARIABLE

Most of the rules for solving a linear inequalities in one variable are similar to those for solving a linear equation in one variable with exception of 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

(a)                  2x – 3 < x + 7

Adding 3 to both sides

2x < x + 10

Subtracting x from both sides

X < 10

(a)                  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

Evaluation

Find the solution set of the inequalities

a) x2 + 5x – 14 < 0

b) 2 – 3x – 9x2> 0

c) 1 – x2 ≤ 0

Quadratic Inequality curve

We recall that th 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

2) 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

b)  9 – x2 ≥ 0

32 – x2 ≥ 0

(3 – x)(3 + x)  ≥  0

3 – x ≥ 0   or  3 + x ≥ 0

– x ≥ -3 or x ≥ -3

x ≤ 3  or  x  ≥ -3

(3 – x) ≤ 0 or (3 + x) ≤ 0

-x  ≤  -3  or  x  ≤  – 3

x ≥ 3 or x ≤ – 3

Solution set  -3  ≤  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

Evaluation

Find the solution set of the inequality

a) │2x – 1│>3

b) │x – 3│ – │x – 1│< 0

c) │x – 3│ ≤│x – 2│

General Evaluation

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

2) For what values of x is 2x2 – 11x + 12 positive?

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

4)The 2nd term of an exponential sequence is 9 while the 4th term is 81. Find the common ratio, the first term and

the  sum of the first  five terms of the sequence.

5) Find the value of the constant k for which the equation 2x2 + (k + 3)x + 2k = 0 has equal roots.

Reading Assignment : F/maths Project 1 pg 104 – 111

WEEKEND ASSIGNMENT

1) Find the range of x for which │2x – 1│> 3

(a) 1 < x < 3/2  (b) -3/2 < x < -1  (c) -3/2 < x < 1 (d) x > 3/2 and x < -1

2) Find the range of the value that satisfies the inequality x2 + 3x – 18 < 0

(a) -3 < x < 6 (b)-3 > x <6 (c)-6 >x >3 (d)-6 >x < 3 (e)-6 < x <3

3) Find the range of values of x for which 2x2 – 5x + 2 ≥ 0

(a) -2<x<-½ (b) ½ <x<2 (c) x < -½ or x ≥ -2 (d) x ≤½ or x ≥ 2

4) Find the range of values of y which satisfies the inequality 2y – 1 < 3 and 2 – y ≤ 5

(a) – 3 ≤ y ≤ 1  (b) – 2 ≤y ≤ 3 (c) -3≤ y ≤ 4 (d) -3 ≤ y ≤ 2

5) Find the range of values of x  for which 1/x + 3 < 2x is satisfy

(a) – 3 < x < 5/2  (b) x < -3 and x > -5/2  (c) x < 1 and x < ½

THEORY

1) Find the range of values of x for which 1 < x2 – x + 1 < 7

2) Find the values of x satisfying |x – 5| – |x – 3| ≥ 0

3) Given that a and b are two real numbers, show that a2 + b2 ≥ 2ab.

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