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Chapter 2 Equations and Inequalities 1 Chapter 2.1 Quadratic Equations 2 Quadratic Equations • General form: ax2 + bx + c = 0, a, b, c are real numbers, a ≠ 0 • General Solution methods: – Factoring; relies on ab=0 means a = 0 or b = 0 – Completing the Square – Quadratic Formula 3 Completing the Square • Work to get the equation into the form: (x – α)2 = β Steps for an equation of the form ax2 + bx = c 1. Divide the equation by a 2. Find b/2, b2 3. (x – b/2) 2 = c + b2/4 4. Take the square root of both sides. 4 Example x2 – 3x + 4 = 0 5 Solution x2 – 3x + 4 = 0 • (x – 3/2) 2 = -4 + 9/4 = 5/4 • x = 3/2 ± 5 /4 6 Another Example • 4x2 – 8x - 16 = 0 7 Solution • 4x2 – 8x - 16 = 0 • x2 – 2x = 4 • (x – 1) 2 = 5 • x=1± 5 8 Quadratic Formula • For ax2 + bx + c = 0, we have −𝑏 ± 𝑏 2 − 4𝑎𝑐 𝑥= 2𝑎 9 Using the Quadratic Formula • x2 = 5 – 3x 10 • x2 = 5 – 3x • Rewrite: x2 + 3x - 5 = 0: a = 1, b = 3, c = 5 𝑥= −𝑏± 𝑏2 −4𝑎𝑐 2𝑎 = −3± 32 −20 2 = (-3 ± 29)/2 11 The Discriminant • We call b2 – 4ac (of −𝑏± 𝑏2 −4𝑎𝑐 ) 2𝑎 the discriminant • It discriminates (determines) what kind of roots we have – If the discriminant is > 0, we have two real roots; if it is > 0 and a perfect square, we have two rational roots – If the discriminant is < 0 we have two complex roots; the roots will be complex conjugates – If the discriminant = 0, we have one rational root (sometimes called a repeated root) 12 Some Practice Use the quadratic formula to determine the type of roots and find the roots • x2 – 6x – 2 = 0 • 16 s2 + 8s + 1 = 0 • (x – 5)(x + 3) = 1 • y2 + y + 1 = 0 13 Solutions • x2 – 6x – 2 = 0; 3 3 2, two reals • 16 s2 + 8s + 1 = 0; s= -1/4, one rational • (x – 5)(x + 3) = 1;1 17; two reals • y2 + y + 1 = 0; [-1 𝑖 3 ] / 2; two complex 14 Using Roots to Build Equations • If we have the roots r and q then we have an equation – r and q can be real or complex – This holds for equations beyond quadratic equations • If the x intercepts of a graph are x = 3 and x = -2, then (x – 3) = 0 and (x + 2) = 0 so (x – 3) (x + 2) = x2 – x – 6 = 0; a quadratic with roots 3 and - 2! 15 We can also use complex roots, radicals • Given roots x = -2 i 5 • (x + 2 - i 5 )(x + 2 + i 5 ) = x2 + 4x +9 = 0 16 Finding Equations Find a quadratic equation with roots as given • 3, 11 • -4, -9 • 2 5 • 1/3[ 4 5 ] 17 Solutions • 3, 11; x2 – 14x + 33 =0 • -4, -9; x2 +13x + 36 = 0 • 2 5; x2 – 4x + 1 = 0 • 1/3[ 4 5 ]; 9x2 – 214x + 11 = 0 18 Examples • Show that (x – p)(x – q) = r2 p and q real and p q has two real roots 19 • (x – p)(x – q) = r2 • x2 – (p + q) x + pq – r2 = 0 • x = [p + q (𝑝 + 𝑞)2 −4𝑝𝑞 + 4 𝑟 2 ] / 2 • Examine quantity under the radical: p2 + q2 + 2pq – 4pq + 4r2 = (p-q) 2 + 4r2 which, as the sum of two squares is positive. • The quantity under the radical is > 0, hence, the roots are real 20 Product and Sum of Roots • If we have x2 + bx + c = 0, and roots r1 and r2, a and b rational, then we have: (r1)(r2) = c and r1 + r2 = - b 21 proof −𝑏+ 𝑏 2 −4𝑎𝑐 • (r1)(r2) = 2𝑎 = −𝑏− 𝑏 2 −4𝑎𝑐 2𝑎 = 𝑏 2 −(𝑏 2 −4𝑐) 4 4𝑐 =c 4 • r1 + r2 = (-2b)/2 = -b 22 Example If we have x + 3x – 7/2 = 0 then our two roots, (r1)(r2) = -7/2 and r1 + r2 = -3 Right now, you can use this to check roots 23 Example • Find A and B such that the roots of x2+ Ax + B = 0 are A and B 24 Solution • Find A and B such that the roots of x2+ Ax + B = 0 are A and B • We need AB = B and A + B = - A • The first equation gives A = 1, the second gives B = -2 • Check: x2+ x - 2 = 0 ; subst x = 1; 1 + 1 – 2 = 0, subst x = -2; 4 – 2 – 2 = 0 25 More Generally • If you have a polynomial with rational coefficients: xn + an-1xn1 + an-2xn-2 …..+ a1x + a0 = 0 • If we factor it as (x – r1)(x- r2)(x – r3)…..(x – rn) = 0, the product of the roots r1 r2 r3 … rn = a0 26 Some more cool facts If a polynomial has rational coefficients: – If ( + i ) is a root, then so is ( - i ) [conjugate root thm] – If a polynomial has an odd degree, than at least one root is real [fundamental thm of algebra] – A polynomial can be factored into terms of the form (x – a) and (x2 + b2) (Note, (x2 + b2) factors to (a + ib)(a – ib) ) [corr. of fundamental thm of algebra] 27 Chapter 2.2 Other Kinds of Equations 28 Absolute Value (Review) • |4| = | -4| = 4 • View absolute value as the “distance from zero” • Solving requires breaking into two equations, one for each branch • BE CAREFUL: Check your results – sometimes one of the roots (solutions) violates the absolute value principles 29 Solve • |2x – 7| = 5 30 Solution • |2x – 7| = 5 • (2x – 7 ) = 5 or (2x – 7) = -5 • x = 6 or x = 1 31 Solve • | x + 1| = 3x - 3 32 Solution • | x + 1| = 3x - 3 • (x + 1) = 3x – 3; x = 2 • Or (x + 1) = 3 – 3x; x = ½; However, x = ½ does not work since 3/2 – 3 < 0; • The only solution is x = 2 33 More Examples • |x – 4| - 5 = 2 • |x + 6| + ½ = 0 • |5 – 6x| = 0 34 Solutions • |x – 4| - 5 = 2; | x – 4| = 8, (x – 4) = 8 or x = 12; (x – 4) = -8, x = -4 • |x + 6| + ½ = 0; |x + 6| = -1/2 no solution • |5 – 6x| = 0; 6x = 5, x = 5/6 35 Moving beyond quadratic equations 36 nth roots • Solve (x – 2)4 = 8 37 • Solve (x – 2)4 = 8 4 • x–2= 8 38 Solve • 3x4 = -64 39 • 3x4 = -64 • No solution; x4 must be > 0, if it is real 40 Solve • 5x5 = -64 41 • 5x5 = -64 • x= - 5 64 5 42 Solving by Factoring • x3 – 3x2 + 2x = 0 43 Solution • x3 – 3x2 + 2x = 0 • x(x2 – 3x + 2) = 0 = x(x – 2)(x – 1) • x = 1, x = 2, x = 0 44 Solve • x4 – 3x2 + 2 = 0 45 • x4 – 3x2 + 2 = 0 • (x2 – 2)(x2 – 1) = 0 • x = 2, x = 1 46 Solve • x4 – 3x2 + 1 = 0 47 Solution • x4 – 3x2 + 1 = 0 • Doesn’t factor! • Can solve for x2 then take roots • Substitution: Let r = x • r = (-3 9 − 4)/2 = (-3 5)/2 • x = (−3 5)/2 48 More examples • 6x -2 – x-1 – 2 = 0 49 Solution • 6x -2 – x-1 – 2 = 0 • Two methods: – Let x-1 = y and solve for y – Multiply both sides of the equation by x2 50 BEWARE THE EXTRANEOUS SOLUTION • If you square both sides of an equation (or any other even power) you may introduce an incorrect (extraneous) solution 𝑥=x–2 squaring gives x = x2 – 4x + 4, or x2 – 5x + 4 = 0 factoring gives (x – 4)(x – 1), but x = 1 does not satisfy the equation if a2 = b2, a may not equal b • Also, if you multiply both sides of an equation by an expression containing the variable you may introduce an additional solution 51 A bit on extraneous roots • Suppose we have x = 2 When we square both sides we get x2 = 4 Solving x2 = 4 gives x = ± 2, only one of which is true in the original equation Squaring the equation has introduced another root • x-1 = 5, or x = 6 Again, squaring gives x2 – 2x + 1 = 25, x2 – 2x - 24= 0 Roots are x = 8 and x = 6, but only x = 6 is true in the original equation Squaring the equation has introduced another root 52 Extraneous roots, continued • Suppose we have x - 2 = 3, clearly x = 5 • Let’s multiply both sides by x x2 – 2x = 3x, or x2 – 5x = 0 • The roots to this equation are x = 0 and x = 5 we introduced an additional root by multiplying by x • The same could hold by multiplying both sides by any function of x • Always check for extra roots when squaring or multiplying by a factor containing a variable 53 Recap We have been working to solve (find the roots of) polynomials • Quadratics: – Taking the square root of both sides – Factoring – Completing the square – Quadratic formula • A quadratic equation has: – Two real roots – Two complex roots (complex conjugates) – One repeated roots The types of roots depend on the discriminant 54 Recap Continued • In a quadratic equation, x2 + bx + c = 0, if b and c are rational, the roots, r1 and r2: – Have a product c: (r1)(r2) = c – Have a sum –b: r1 + r2 = -b • We expanded these techniques to solve equations of higher degree, e.g., cubics and quartics: – Taking roots of both sides – Factoring – Substitution to get quadratic equations • Reminder: If we square both sides, or multiply by a variable, we may introduce extraneous roots – check them. 55 Equations with Radicals • Solve: 2 + 10 − 𝑥 = x 56 Solution • Solve: 2 + 10 − 𝑥 = x • 10 − 𝑥 = x - 2, square both sides • 10 – x = x2 - 4x + 4 or x2 - 3x – 6 = 0 • x = [3 ± 9 + 24 ]/2 = [3± 33 ]/2 • x = [3 ± 33 ]/2; is not problem with either root for a negative discriminant as 33 < 6 • Try in x2 - 3x – 6 = 0 x2 = [9 ± 6 33 + 33]/4 = [42 ± 6 33]/4 -3x = -[9 ± 3 33]/2 = -[18 ± 6 33]/4 • x2 - 3x – 6 =[42 -18]/4 – 6 = 24/4 – 6 = 0 57 In General • If you have an equation with a radical, isolate the radical and square both sides • What if we have two radicals? Keep isolating and squaring 58 Example • Solve 2𝑥 + 3 −2 𝑥 − 2= 1 59 Solution • 2𝑥 + 3 −2 𝑥 − 2= 1 • 2𝑥 + 3 = 1 + 2 𝑥 − 2 or 2x + 3 = 1 + 4x – 8 + 4 𝑥 − 2 • -2x + 10 = 4 𝑥 − 2 which is –x + 5 = 2 𝑥 − 2 • Square both sides: x2 – 10X + 25 = 4x – 8 (x – 3)(x – 11) = 0, x = 3, x = 11 only x = 3 works 60 Chapter 2.3 Inequalities 61 Inequalities (Review) • Treat just like an equation to solve them: add the same thing to both sides, multiply both sides by the same thing, EXCEPT: when you multiply or divide by a negative number, you must change the orientation of the inequality • 2x – 3 < 5 gives 2x < 8 or x < 4 • However, -2x – 3 < 5 gives -2x < 8 or x > - 4 (You can do this without multiplying: -3 – 5 < 2x, through addition, then x > - 4) 62 Solution Set Notation • If the solution is t > 4, we write (4, ∞), sometimes t (4, ∞) If the solution is t ≥ 4, we write [4, ∞) • Sometimes we get a solution like -1 < t < 4, which we write as (-1, 4) • If we have |t| > 4, we have the two sets (-∞, -4) or (4, ∞) We can write the or as the union of the sets, (-∞, -4) (4, ∞) 63 Compound Inequality We have two or more inequalities we have to solve: 3 −𝑥 -½ < <½ 4 Solve them separately: -1/2 < 3 −𝑥 4 3 −𝑥 < 4 1/2 2 > (3 – x) (3 – x) > -2 x>1 5>x Then combine to get the solution: 1<x<5 64 Pitfalls When you complete both parts, you might find: – There is no joint solution: x < 2 and x > 3 has no solution – Only one part of the solution is relevant: x < 2 and x < 5, leaving x < 2 as the solution Just as with an equation, you can put your answer into the inequality to check it 65 Examples • 1 – 2(t + 3) – t < 1 – 2t • (x – 1)/4 – (2x + 3) /5 < x • -3 < 2x + 1 < 5 • 2/3 < (5 – 3t)/2 < ¾ • 9/10 < (3x – 1) / (-2) < 91/100 66 Solutions • 1 – 2(t + 3) – t < 1 – 2t; 1 – 2t – 6 – t < 1 – 2t -3t -5 < 1 – 2t, t > -6 • (x – 1)/4 – (2x + 3) /5 < x 5x – 5 – 8x – 12 < 20x , 23x > -17, x > -17/23 • -3 < 2x + 1 < 5 2x > -4, x > -2; 2x < 4, x < 2; -2 < x < 2 • 2/3 < (5 – 3t)/(-2) < ¾; 8 < -30 + 18t < 9 18t > 38 or t > 19/9, 18t < 39, t < 13/6, 19/9 < t < 13/6 • 9/10 < (3x – 1) / (-2) < 91/100 90< 50 – 150x < 91; -40/150 > x, -41/150 > x, so we have -45/15 > x > -41/150 67 Inequalities with Absolute Values • |x| < 1 becomes -1 < x < 1; the distance from x to zero is < 1 • |x| > 1 becomes x < -1 or x > 1; the distance is > 1 68 Example • |x – 5| < 3 69 Solution • |x – 5| < 3 • x – 5 < 3 or -x + 5 < 3 x < 8 or x > 2; 2<x<8 70 Example • |1 – t/5| > 3 71 Solution • |1 – t/5| > 3 • 1 – t/5 > 3 or 1-t/5 < -3 • So t < -10 or t > 20 72 Examples • |x – 4| > 4 • |x – 4| < 4 • 3(𝑥 −2) 4(𝑥 −1) + 4 3 <2 • | 3(x+2)2 – 3x2| < 1/10 73 • |x – 4| > 4; x – 4 > 4 or x -4 < -4, x > 8 or x < 0 • |x – 4| < 4; x – 4 < 4 and x – 4>-4, x < 8 and x > 0; 0 < x < 8 • 3(𝑥 −2) 4(𝑥 −1) + 4 3 < 2; |9x – 18 + 16x – 16| < 24 |25x – 34| < 24, or 25x – 34 > 24 and 25x – 34 > -24 x < 58/25, x > 2/5, 2/5 < x < 58/25 • | 3(x+2)2 – 3x2| < 1/10 this is |(x + 2) 2 – x2| < 1/30 |4x + 4| < 1/30, 4x + 4 < 1/30 and 4x + 4 > -1/30 -121/120 < x < -119/120 74 Chapter 2.4 More on Inequalities 75 Key Numbers for Graphics • Graph y = x2 – 4x + 3 • The graph has 3 regions: x < 1, 1 < x < 3, and x > 3; at these points y changes sign 76 We can use this to solve inequalities! • y > x2 – 4x + 3 or y < x2 – 4x + 3 • Clearly, from the graph, the first holds for x < 1 and x > 3, the second holds for 1 < x < 3 • However, we can solve too! • x2 – 4x + 3 = (x – 3)(x – 1) For x < 1, both terms are < 0, so y > 0 For x > 3, both terms are >0, so y > 0 For x between 1 and 3, one is positive and one negative, so y < 0 77 Example • Find the regions where x3 – 2x2 – 3x > 0 78 Solution • x3 – 2x2 – 3x > 0 • Factor: x (x – 3)(x + 1) Roots are x = 0, x = 3, x = -1. Put in order: -1, 0, 3 • For x < -1, all terms are negative, so the expression is negative • For -1<x<0, two terms are negative, so the expression is positive • For 0 < x < 3, two terms are positive, so the expression is negative • For x > 3, all terms are positive, so the expression is positive 79 Example • x4 < 14x3 – 48x2 80 Solution • x4 < 14x3 – 48x2 • Divide by x; we can do that unless x = 0. However, zero is also a critical point; • x2 < 14x – 48 • Rewrite: x2 – 14x + 48 = 0, (x – 6)(x – 8) = 0, x = 6, x = 8, x = 0 are the critical or key points • Put numbers into the inequality to check regions: For 0< x < 6: Let x = 1. 1 ? 14 – 48, clearly 1 > 14-48, so the inequality doesn’t hold for x < 6 For x > 8: Let x = 10. 100 ? 140 – 48, clearly 100 > 140-48, so the inequality doesn’t hold for x > 8 Try x = 7: 49 ? 98 – 48 = 50. 49 < 50, so holds for 6 < x < 8 • We also need to check x < 0, since 0 is also a root. Use – 1, - 1 ? -14 – 48,81 Does not holds for x < 0 There aren’t always key numbers • x2 – 4x + 5 > 0 • If we make it an equation, it has no real solutions. But, if it has no real solutions, it doesn’t intersect the x axis and can’t change sign. • Therefore, it is always > 0 or < 0 • Substituted x = 0 shows that it is always positive, or always true 82 Quotients • 𝑥+3 𝑥 −4 >0 83 Solution • 𝑥+3 𝑥 −4 >0 • Key numbers are -3 and 4 (and x cannot be 4!) • -4 < -3, so try it. We get -1 / -8 which is > 0 • 0 is between -3 and r, 3/-4 < 0, so doesn’t hold • 5 > 4, and 8/1 > 0, so holds • Equation holds for x < -3 or x > 4 84 What if we have ≥ instead of > • y ≥ x2 – 4x + 3 • x3 – 2x2 – 3x ≥ 0 • x2 – 4x + 5 ≥ 0 • 𝑥+3 𝑥 −4 ≥0 • Still try key points 85 Solution • y ≥ x2 – 4x + 3; points are 1 and 3, 1 works, 3 too • x3 – 2x2 – 3x ≥ 0; points are 0, -1, 3; all solve equation • x2 – 4x + 5 ≥ 0; no real number solves the equation, but the inequality holds for all real numbers • 𝑥+3 𝑥 −4 ≥ 0; can’t have x = 4, but x = -3 solves 86 What you need to know • Working with Complex Numbers • Solving equations using substitution, factoring quadratic formula • Solving equations with radicals • Checking for extraneous solutions • Solving inequalities (review) • Key Numbers 87