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Chapter 3 β 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
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03 Algebra III Surd Equations πΊππππ π+π + π+π =π
πΊππππ π+π + π+π =π As we have more than one surd term, we leave one surd term on one side ofβ―theβ―equationβ― andβ―moveβ―everyβ―otherβ― termβ―ontoβ―theβ―otherβ―side, toβ―simplify the arithmetic.β― π₯+7 =5β π₯+2 π₯ = 5β π₯+2 2 Squaring both sides π₯ + 7=(5β π₯+2 )(5β π₯+2 ) π₯+7=25β5 π₯+2 β5 π₯+2 +( π₯+2 ) 2 π₯+7=25β10 π₯ π₯+2 π₯+7=27β10 π₯ π₯ Again, isolate the surd term on one side of the equation. Check (in the original equation) πΌπ π₯=2 10( π₯+2 ) = 20 πΏπ»π= =3+2=5 π₯+2 = 2 Squaring both sides π
π»π=5 π₯+2=4 π₯=2 Which is true. π₯=2
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03 Algebra III Linear Inequalities 1 2 3 β1 β2 β3 β4 β3.5
Solve the inequality and show the solution set on the numberline: Multiply every term by 3 β1β€ 2π₯+4 3 <2 β3β€2π₯+4<6 β3β4β€2π₯<6β4 β7β€2π₯<2 β7 2 β€π₯<1 1 2 3 β1 β2 β3 β4 β3.5
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Sketch the quadratic function.
03 Algebra III Quadratic and Rational Inequalities QUADRATIC INEQUALITY RATIONAL INEQUALITY Solve ππ+π π+π <π, πβπΉ, πβ βπ Solve π2 + 7π + 12 β€ 0, x β R. Let π₯2 + 7π₯ + 12 = 0 and solve. Multiply both sides by (π₯+1) 2 π₯2 + 7π₯ + 12 = 0 π₯+1 2 (2π₯+4) π₯+1 <3 (π₯+1) 2 (π₯ + 3)(π₯ + 4) = 0 π₯ = β3 OR π₯ = β 4 (π₯+1)(2π₯+4)<3 (π₯+1) 2 Sketch the quadratic function. 2 π₯ 2 +6π₯+4<3 π₯ 2 +6π₯+3 0< π₯ 2 β1 π₯ 2 β1>0 Sketch the quadratic function π₯ 2 β1. As the function is > 0 the solution lies above the x-axis). As the function is β€ 0 the solution lies on or below the x-axis. The solution is π₯ < β1 OR π₯ > 1 The solution is β4β€π₯β€β3
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Solve for x if 3|x + 1| β |x + 5| = 0.
03 Algebra III Absolute Value (Modulus) The Modulus of a number is the non-negative value of the number. Example |β7| = 7 Geometrically, the absolute value is how far away the number is from zero on the numberline. Solve for x if 3|x + 1| β |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| π π₯ =3|π₯+1| (1,6) Squaring both sides removes the modulus notation. (β2,3) 9(x + 1)2 = (x + 5)2 π π₯ =|π₯+5| This simplifies to x2 + x β 2 = 0 Solving gives The two functions intersect at (β2,3) and (1,6). x = β2 OR x = 1 β΄ x = β2 OR x = 1
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Solve the inequality |x + 1| > 2|x + 3|, x β R.
03 Algebra III Modulus Inequalities Solve the inequality |x + 1| > 2|x + 3|, x β 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 π π₯ =2|π₯+3| 3x2 + 22x + 35 < 0 π π₯ =|π₯+1| Let 3x2 + 22x + 35 = 0. (3x + 7)(x + 5) = 0 β΄ x = β OR x = β5 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 β5<π₯<β 7 3 The solution is β5<π₯<β 7 3
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03 Algebra III Inequalities: Proofs
Prove that if a and b are real numbers, then a2 + b2 β₯ 2ab. a2 + b2 β₯ 2ab β a2 + b2 β 2ab β₯ 0 β (a β b)2 β₯ 0. True, as (a β b) β R and (real)2 β₯ 0 β΄ a2 + b2 β₯ 2ab If a, b β R, prove that a2 + 4b2 β 10a + 25 β₯ 0. a2 β 10a b2 β₯ 0 (a β 5)(a β 5) + (2b)2 β₯ 0 β (a β 5)2 + (2b)2 β₯ 0 True, as a, b β R, and (real)2 + (real)2 β₯ 0.
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03 Algebra III Discriminants b2 β 4ac > 0 b2 β 4ac = 0
When using the formula π₯= βπΒ± π 2 β4ππ 2π b2 β 4ac is called the discriminant. The value of the discriminant, b2 β 4ac , can be used to determine whether the graph of the function (the parabola) cuts, touches or does not cut the x-axis. b2 β 4ac > 0 b2 β 4ac = 0 b2 β 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 β 4ac = 0 a = 1, b = 6, c = k (6)2 β 4(1)(k) = 0 k = 9
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