Inequalities – Learning Outcomes pg 120-136 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 >)

Use Graphic Strategies pg 120-126 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.

Use Algebraic Strategies pg 120-128 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

Use Algebraic Strategies pg 120-128 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

Use Algebraic Strategies pg 120-128 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

Use Algebraic Strategies pg 120-128 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.

Use Algebraic Strategies pg 120-128 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 −2 2 3 ⇒𝑥=1.215 or 𝑥=−0.549 Reform this into factorised form: ⇒ 𝑥−1.215 𝑥+0.549 ≤0 And work out what values of 𝑥 will make the product negative.

Use Algebraic Strategies pg 120-128 Use Algebraic Strategies ⇒ 𝑥−1.215 𝑥+0.549 ≤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.

Use Algebraic Strategies pg 120-128 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

Use Algebraic Strategies pg 120-128 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.

Use Algebraic Strategies pg 120-128 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.

Use Algebraic Strategies pg 120-128 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

Use Algebraic Strategies pg 120-128 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.

pg 128-131 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

pg 128-131 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

Use Modulus Solve 𝑥+3 =2 Solve 2𝑥+3 =7 Solve 2𝑥−1 = 𝑥+2 pg 128-131 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

Solve Modulus Inequalities pg 132-134 Solve Modulus Inequalities For modulus inequalities, the direction of the inequality determines how to solve it.

Solve Modulus Inequalities pg 132-134 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

Solve Modulus Inequalities pg 132-134 Solve Modulus Inequalities Solve 𝑥−2 >1 ⇒𝑥−2>1 ⇒𝑥>3 or ⇒𝑥−2<−1 ⇒𝑥<1

Solve Modulus Inequalities pg 132-134 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.