1. Inequalities – Learning Outcomespg
Inequalities – Learning Outcomes
Use graphic, numeric, algebraic, and mental strategies to solve inequalities of the form:
𝑎 𝑥 2 +𝑏𝑥+𝑐≤𝑘 (or ≥, <, >)
𝑎𝑥+𝑏 𝑐𝑥+𝑑 ≤𝑘 (or ≥, <, >)
Use modulus notation
Solve inequalities of the form 𝑥−𝑎 <𝑏 (or >)

2. Use Graphic Strategiespg
Use Graphic Strategies
Given the function 𝑓 𝑥 =3𝑥−2,
Plot a graph of 𝑓 𝑥 in the domain −3≤𝑥≤3.
Use your graph to solve the inequality 𝑓(𝑥)≤0.
Given the function 𝑔 𝑥 = 4𝑥 3 +1,
Plot a graph of 𝑔(𝑥) in the domain −6≤𝑥≤−2.
Use your graph to solve the inequality 𝑔 𝑥 <0.
Given the function ℎ 𝑥 = 𝑥 2 +2𝑥−3,
Plot a graph of ℎ(𝑥) in the domain −4≤𝑥≤2.
Use your graph to solve the inequality ℎ 𝑥 >0.

3. Use Algebraic Strategiespg
Use Algebraic Strategies
The rules for solving inequalities are almost the same as the rules for solving equalities.
𝟑<𝟔
3+3<6+3⇒6<9
Addition valid
3−3<6−3⇒0<3
Subtraction valid
3×3<6×3⇒9<18
Multiplication valid
3 3 < 6 3 ⇒1<2
Division valid
3×−3<6×−3⇒−9≮−18
Negative multiplication invalid
3 −3 < 6 −3 ⇒−1≮−2
Negative division invalid

4. Use Algebraic Strategiespg
Use Algebraic Strategies
Multiplying or dividing by a negative number makes the inequality invalid.
When multiplying or dividing by a negative number, the direction of the inequality must be reversed.
e.g. 3×−3<6×−3⇒−9>−18

5. Use Algebraic Strategiespg
Use Algebraic Strategies
Solve 3𝑥−2≤0
⇒3𝑥≤2
⇒𝑥≤ 2 3
Solve 4𝑥 3 +1<0
⇒ 4𝑥 3 <−1
⇒4𝑥<−3
⇒𝑥<− 3 4

6. Use Algebraic Strategiespg
Use Algebraic Strategies
Solve 𝑥 2 +2𝑥−3>0
⇒ 𝑥+3 𝑥−1 >0
When is the product greater than zero?
Consider where each factor is positive and negative by looking at the equivalent root. −𝟑 𝟏
(𝑥+3)
(𝑥−1)
(𝑥+3)(𝑥−1)
>0
<0
So 𝑥<−3 and 𝑥>1 are valid solutions for 𝑥.

7. Use Algebraic Strategiespg
Use Algebraic Strategies
Solve 3 𝑥 2 −2𝑥−2≤0
Not factorisable – pretend it’s an equation and use quadratic formula:
𝑥= − −2 ± −2 2 −4 3 −
⇒𝑥=1.215 or 𝑥=−0.549
Reform this into factorised form:
⇒ 𝑥− 𝑥 ≤0
And work out what values of 𝑥 will make the product negative.

8. Use Algebraic Strategiespg
Use Algebraic Strategies
⇒ 𝑥− 𝑥 ≤0
And work out what values of 𝑥 will make the product negative.
−𝟎.𝟓𝟒𝟗
𝟏.𝟐𝟏𝟓
(𝑥+0.549)
(𝑥−1.215)
(𝑥−1.215)(𝑥+0.549) ≥0 ≤0
So −0.549≤𝑥≤1.215 is the solution set for 𝑥.

9. Use Algebraic Strategiespg
Use Algebraic Strategies
Determine the values of 𝑘∈ℝ for which the quadratic equation 𝑥 2 −𝑘𝑥+ 3𝑘−8 =0 has real roots.
Remember the quadratic formula:
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
will yield real roots when 𝑏 2 −4𝑎𝑐 is positive or zero (i.e. 𝑏 2 −4𝑎𝑐≥0)
For this quadratic, −𝑘 2 −4 1 3𝑘−8 ≥0
⇒ 𝑘 2 −12𝑘+32≥0

10. Use Algebraic Strategiespg
Use Algebraic Strategies
𝑘 2 −12𝑘+32≥0
⇒ 𝑘−4 𝑘−8 ≥0 𝟒 𝟖
(𝑘−4)
(𝑘−8)
(𝑘−4)(𝑘−8) ≥0 ≤0
So 𝑘≤4 and 𝑘≥8 are valid values for 𝑘.

11. Use Algebraic Strategiespg
Use Algebraic Strategies
Solve 2𝑥+3 𝑥−1 <1 for 𝑥∈ℝ, 𝑥≠1.
(why is it important that 𝑥≠1?)
If this were an equation, the first step we would normally take is to multiply both sides by 𝑥−1 to get rid of the fraction.
Given that this is an inequality, remember that multiplying by a negative number reverses the inequality, while multiplying by a positive number preserves the inequality.

12. Use Algebraic Strategiespg
Use Algebraic Strategies
To be sure of our answer, we must be sure to multiply by a positive number.
𝑥−1 2 will still eliminate the fraction, while we guarantee that it is positive:
𝑥−1 2 × 2𝑥+3 𝑥−1 <1× 𝑥−1 2
⇒ 2𝑥+3 𝑥−1 <1 𝑥−1 𝑥−1
⇒2 𝑥 2 −2𝑥+3𝑥−3< 𝑥 2 −𝑥−𝑥+1
⇒2 𝑥 2 +𝑥−3< 𝑥 2 −2𝑥+1
⇒ 𝑥 2 +3𝑥−4<0

13. Use Algebraic Strategiespg
Use Algebraic Strategies
𝑥 2 +3𝑥−4<0
⇒ 𝑥+4 𝑥−1 <0 −𝟒 𝟏
(𝑥+4)
(𝑥−1)
(𝑥+4)(𝑥−1)
>0
<0
So −4<𝑥<1 is the valid solution set.

14. pg
Use Modulus
The modulus of a number, |𝑥|, outputs the positive version of that number.
e.g. 5 =5
e.g. −5 =5
If given |𝑥|, there is no way to tell whether 𝑥 is positive or negative.
If 𝑥 =𝑎, then there are two possibilities:
𝑥=𝑎, or
−𝑥=𝑎

15. pg
Use Modulus
If 𝑥 =|𝑦|, since we cannot tell the sign of 𝑥 or 𝑦, we get two possibilities:
𝑥=𝑦, or
𝑥=−𝑦
Finally, since squaring a number ensures a positive result, 𝑥 2 = 𝑥 2

16. Use Modulus Solve 𝑥+3 =2 Solve 2𝑥+3 =7 Solve 2𝑥−1 = 𝑥+2pg
Use Modulus
Solve 𝑥+3 =2
Solve 2𝑥+3 =7
Solve 2𝑥−1 = 𝑥+2
Solve 3𝑥−5 = 7𝑥+1
Solve 𝑚+1 = 𝑚 2 +5

17. Solve Modulus Inequalitiespg
Solve Modulus Inequalities
For modulus inequalities, the direction of the inequality determines how to solve it.

18. Solve Modulus Inequalitiespg
Solve Modulus Inequalities
For 𝑥 <𝑎, the solution is given by −𝑎<𝑥<𝑎
For 𝑥 >𝑎, the solution is given by −𝑎>𝑥 or 𝑥>𝑎
Solve 𝑥−2 <5
⇒−5<𝑥−2<5
⇒−5+2<𝑥−2+2<5+2
⇒−3<𝑥<7

19. Solve Modulus Inequalitiespg
Solve Modulus Inequalities
Solve 𝑥−2 >1
⇒𝑥−2>1
⇒𝑥>3 or
⇒𝑥−2<−1
⇒𝑥<1

20. Solve Modulus Inequalitiespg
Solve Modulus Inequalities
Solve 1< 𝑥−2 <5
From previous questions, we know:
1< 𝑥−2 ⇒𝑥<1 or 𝑥>3
𝑥−2 <5⇒−3<𝑥<7
Combining the possible values for 𝑥 yields:
−3<𝑥<1 or 3<𝑥<7 as the solution sets.