Chapter 3 – Algebra III 03 Learning Outcomes In this chapter you have learned to: Solve equations containing surds Find solutions to inequalities of the following forms: linear, quadratic and rational Understand absolute value (modulus) and use its notation, |x| Find solutions to modulus inequalities Use discriminants to determine the nature of roots of quadratic equations Prove algebraic inequalities

03 Algebra III Surd Equations 𝑺𝒐𝒍𝒗𝒆 𝒙+𝟕 + 𝒙+𝟐 =𝟓 𝑺𝒐𝒍𝒗𝒆 𝒙+𝟕 + 𝒙+𝟐 =𝟓 As we have more than one surd term, we leave one surd term on one side of the equation  and move every other  term onto the other side, to simplify the arithmetic.  𝑥+7 =5− 𝑥+2 𝑥+7 2 = 5− 𝑥+2 2 Squaring both sides 𝑥 + 7=(5− 𝑥+2 )(5− 𝑥+2 ) 𝑥+7=25−5 𝑥+2 −5 𝑥+2 +( 𝑥+2 ) 2 𝑥+7=25−10 𝑥+2 + 𝑥+2 𝑥+7=27−10 𝑥+2 + 𝑥 Again, isolate the surd term on one side of the equation. Check (in the original equation) 𝐼𝑓 𝑥=2 10( 𝑥+2 ) = 20 𝐿𝐻𝑆= 2+7 + 2+2 =3+2=5 𝑥+2 = 2 Squaring both sides 𝑅𝐻𝑆=5 𝑥+2=4 𝑥=2 Which is true. 𝑥=2

03 Algebra III Linear Inequalities 1 2 3 −1 −2 −3 −4 −3.5 Solve the inequality and show the solution set on the numberline: Multiply every term by 3 −1≤ 2𝑥+4 3 <2 −3≤2𝑥+4<6 −3−4≤2𝑥<6−4 −7≤2𝑥<2 −7 2 ≤𝑥<1 1 2 3 −1 −2 −3 −4 −3.5

Sketch the quadratic function. 03 Algebra III Quadratic and Rational Inequalities QUADRATIC INEQUALITY RATIONAL INEQUALITY Solve 𝟐𝒙+𝟒 𝒙+𝟏 <𝟑, 𝒙∈𝑹, 𝒙≠−𝟏 Solve 𝒙2 + 7𝒙 + 12 ≤ 0, x ∈ R. Let 𝑥2 + 7𝑥 + 12 = 0 and solve. Multiply both sides by (𝑥+1) 2 𝑥2 + 7𝑥 + 12 = 0 𝑥+1 2 (2𝑥+4) 𝑥+1 <3 (𝑥+1) 2 (𝑥 + 3)(𝑥 + 4) = 0 𝑥 = –3 OR 𝑥 = – 4 (𝑥+1)(2𝑥+4)<3 (𝑥+1) 2 Sketch the quadratic function. 2 𝑥 2 +6𝑥+4<3 𝑥 2 +6𝑥+3 0< 𝑥 2 −1 𝑥 2 −1>0 Sketch the quadratic function 𝑥 2 −1. As the function is > 0 the solution lies above the x-axis). As the function is ≤ 0 the solution lies on or below the x-axis. The solution is 𝑥 < –1 OR 𝑥 > 1 The solution is −4≤𝑥≤−3

Solve for x if 3|x + 1| – |x + 5| = 0. 03 Algebra III Absolute Value (Modulus) The Modulus of a number is the non-negative value of the number. Example |−7| = 7 Geometrically, the absolute value is how far away the number is from zero on the numberline. Solve for x if 3|x + 1| – |x + 5| = 0. Method 1 (Algebra) Method 2 (Graph) Leave one modulus term on one side And the other terms on the other side. We plot f(x) = 3|x + 1| and g(x) = |x + 5|. -12 -10 -8 -6 -4 -2 2 -1 1 3 4 5 6 7 8 x y 3|x + 1| = |x + 5| 𝑓 𝑥 =3|𝑥+1| (1,6) Squaring both sides removes the modulus notation. (−2,3) 9(x + 1)2 = (x + 5)2 𝑔 𝑥 =|𝑥+5| This simplifies to x2 + x – 2 = 0 Solving gives The two functions intersect at (–2,3) and (1,6). x = –2 OR x = 1 ∴ x = –2 OR x = 1

Solve the inequality |x + 1| > 2|x + 3|, x ∈ R. 03 Algebra III Modulus Inequalities Solve the inequality |x + 1| > 2|x + 3|, x ∈ R. Method 1 (Algebra) Method 2 (Graph) |x + 1|2 > 4|x + 3|2 We plot f(x) = |x + 1| and g(x) = 2|x + 3|. x2 + 2x + 1 > 4(x2 + 6x + 9) x2 + 2x + 1 > 4x2 + 24x + 36 -12 -10 -8 -6 -4 -2 2 -1 1 3 4 5 6 7 8 x y 𝑔 𝑥 =2|𝑥+3| 3x2 + 22x + 35 < 0 𝑓 𝑥 =|𝑥+1| Let 3x2 + 22x + 35 = 0. (3x + 7)(x + 5) = 0 ∴ x = − 7 3 OR x = –5 As f(x) > g(x) As the function is < 0 the solution lies below the x-axis. As f(x) > g(x) the solution is the part of the Graph where f(x) lies above g(x). The solution is −5<𝑥<− 7 3 The solution is −5<𝑥<− 7 3

03 Algebra III Inequalities: Proofs Prove that if a and b are real numbers, then a2 + b2 ≥ 2ab. a2 + b2 ≥ 2ab ⇒ a2 + b2 – 2ab ≥ 0 ⇒ (a – b)2 ≥ 0. True, as (a – b) ∈ R and (real)2 ≥ 0 ∴ a2 + b2 ≥ 2ab If a, b ∈ R, prove that a2 + 4b2 – 10a + 25 ≥ 0. a2 – 10a + 25 + 4b2 ≥ 0 (a – 5)(a – 5) + (2b)2 ≥ 0 ⇒ (a – 5)2 + (2b)2 ≥ 0 True, as a, b ∈ R, and (real)2 + (real)2 ≥ 0.

03 Algebra III Discriminants b2 – 4ac > 0 b2 – 4ac = 0 When using the formula 𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 b2 – 4ac is called the discriminant. The value of the discriminant, b2 – 4ac , can be used to determine whether the graph of the function (the parabola) cuts, touches or does not cut the x-axis. b2 – 4ac > 0 b2 – 4ac = 0 b2 – 4ac < 0 Real Roots Equal Roots No Real Roots Find the value of k if x2 + 6x + k has two equal roots. Equal roots: b2 – 4ac = 0 a = 1, b = 6, c = k (6)2 – 4(1)(k) = 0 k = 9