Notation of inequality includes:
“ < ” notation less than
“ > ” notation more than
“ ” notation less than or equal to
“ ” notation more than or equal to
Solution of an inequality of one variable in the form of intervals or intervals that can be drawn on a line number
While a linear inequality of one variable is an inequality that contains one variable with the highest power of one.
There are four terms in the interval, namely open interval, closed interval, finite interval and infinite interval.
For more details, follow the following figure for the x variable:
Another form of inequality notation is the not equal sign (written ) However, in the discussion of this chapter, this notation is not described in depth.
An inequality notation can change due to certain operations. These changes can be explained in the following inequalities:
Characteristics of inequalities:
(1) The sign/notation of an inequality does not change if the addition or subtraction of the same number (variable) is performed on both sides of the inequality
Example: 3 < 6
3 + 4 < 6 + 4 (both sides are added 4)
7 < 10
(2) The sign/notation of an inequality does not change if the multiplication or division of the same positive number (variable) is performed on both sides of the inequality
Example: 3 < 6
3 x 2 < 6 x 2 (both sides are multiplied by 2)
6 < 12
(3) The sign/notation of an inequality will change if the multiplication or division of the same negative number (variable) is performed on both sides of the inequality
Example: 3 .
_ _ _ _ 3x – 6 < 12 (b) 5x + 3 3x – 7 Answer (a) 3x – 6 < 12 3x < 12 + 6 3x < 18 x < 6 (b) 5x + 3 3x – 7 5x – 3x – 7 – 3
2x –10
x –5
02. Find the solution interval for the following inequality:
(a) 4x – 6 < 9x – 21 (b) 3x – 5 7x + 11
Answer
(a) 4x – 6 < 9x – 21 (b ) 3x – 5 7x + 11
4x – 9x < 6 – 21 3x – 7x 5 + 11
–5x < –15 –4x 16
x > 3 x –4
03. Find the solution interval for the following inequality:
(a) –8 < 3x + 4 < 22 (b) –3
Answer
(a) – 8 < 3x + 4 < 22
–8 – 4 < 3x + 4 – 4 < 22 – 4
–12 < 3x < 18
–4 < x < 6
(b) –3 9 – 4x 29
–3 – 9 9 – 4x – 9 29 – 9
–12 –4x 20
3 x –5
–5 x 3
General form of quadratic inequality :
ax 2 + bx + c < 0
ax 2 + bx + c > 0
ax 2 + bx + c 0
ax 2 + bx + c 0
The solution to the inequality is a finite interval or infinite interval with the following rules:
If p and q are the roots of the equation ax 2 + bx + c = 0, then p and q are the boundaries of the solution interval for the quadratic inequality.
If D = b 2 – 4ac is the discriminant, then the solution of the quadratic inequality can be explained as follows:
For positive discriminant (D > 0), there will be two interval boundary points, namely p and q so that the solution of the quadratic inequality can be assisted by sketching the graph of the function the following square
For zero discriminant (D = 0), then there will be one interval limit point, for example p (p = q) so that solving quadratic inequalities can be assisted by the following graph sketch of quadratic function
ax 2 + bx + c <0 the solution is p <x <p
a x 2 + bx + c < 0 solution x < p or x > p
or x satisfies all real numbers except p
a x 2 + bx + c 0 solution x p or x p
or x satisfies all real numbers
a x 2 + bx + c > 0 the solution is p < x < p
or there is no x value that satisfies
a x 2 + bx + c 0 the solution is p ≤ x p
or x = p
For negative discriminant (D < 0), then no there is an interval limit point, so that solving quadratic inequalities can be assisted by the following graph sketch of the quadratic function
a x 2 + bx + c <0 the solution has no value of x that satisfies
a x 2 + bx + c ≤ 0 the solution has no value of x that satisfies
a x 2 + bx + c> 0 the solution satisfies all real numbers x
a x 2 + bx + c ≥ 0 the solution satisfies all real numbers x
a x 2 + bx + c < 0 the solution satisfies all real numbers x
a x 2 + bx + c 0 the solution satisfies all real numbers x
a x 2 + bx + c > 0 the solution is no x value satisfies
a x 2 + bx + c 0 solution there is no value of x that satisfies
The steps for solving the inequality are as follows:
(1) Change the right side of the inequality to 0
(2) Determine the limits of the interval, namely the roots of the quadratic equation
(3) State on a number line or graph it
(4) Determine the completion interval
For more details, it will be described in the following example problem:
01. Determine the interval for solving the following inequalities:
(a) x 2 – x – 12 < 0 (b) x 2 – 9 0
(c) –3 x 2 + 9x + 30 > 0 (d) 10x – x 2 24
Jawab
(a) x2 – x – 12 < 0
(x + 3)(x – 4) < 0
x = –3 dan x = 4
–3 < x < 4
(b) x2 – 9 ≥ 0
(x + 3)(x – 3) ≥ 0
x = –3 dan x = 3
x ≤ –3 atau x ≥ 3
(c) –3x2 + 9x + 30 > 0
x2 – 3x – 10 < 0
(x + 2)(x – 5) < 0
x = –2 dan x = 5
–2 < x < 5
(d) x2 – x – 12 < 0
(x + 3)(x – 4) < 0
x1 = –3 dan x2 = 4
–3 < x < 4
02. Find the solution interval for the following inequalities:
(a) x 2 – 2x + 8 > 0 (b) 15x – x 2 – 18 x 2 + 3x
Answer
(a) x 2 – 2x + 8 > 0
D = (– 2)2 – 4(1)(8)
D = –28 < 0
No interval limit
So x satisfies all real numbers
(b) 15x – x 2 – 18 x 2 + 3x
15x – x2 – 18 – x 2 – 3x 0
–2 x 2 – 12x – 18 0
x2 + 6x + 9 0
(x + 3)(x + 3) 0
x = –3
–3 x –3
Or a value that satisfies only for x = –3
03. Determine the interval for solving the following inequality:
( a) x 2 – 8x + 16 > 0 (b) x 2 + 10x + 25 < 0
Answer
(a) x 2 – 8x + 16 > 0
(x – 4)(x – 4) > 0
x = 4
x < 4 or x > 4
Or the value of x satisfies for all real numbers except 4
(b) x 2 + 10x + 25 < 0
(x + 5)(x + 5) > 0
x = –5
–5 < x < –5
Or no value of x satisfies
4. A shoe company manufactures and sells various models of shoes. For one particular shoe model is estimated to be sold for a rupiah. If in one week it costs M rupiah and the income received is P rupiah and it is formulated M = 2,000,000 – 40,000a and P = 20,000a – 400a 2 then what is the price limit for the unit shoes that must be sold in order for the company to make a profit?
Answer
In order to make a profit then:
P > M
20000a – 400a 2 > 2000000 – 40000a
20000a – 400a 2 – 2000000 + 40000a > 0
–400a2 + 60000a – 2000000 > 0
a 2 – 150a + 5000 < 0
(a – 100)(a – 50) < 0
Limit interval a1 = 100 and a 2 = 50
So the interval for the price of shoes is: 50 < a < 100