# how to use barcode in c#.net Worked-Out Solutions to Exercises: s 21 to 29 in Software Generating QR Code 2d barcode in Software Worked-Out Solutions to Exercises: s 21 to 29

692 Worked-Out Solutions to Exercises: s 21 to 29
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Because d = 0, we know that the quadratic we get by setting the trinomial equal to 0 has one real root with multiplicity 2 That means the original cubic has one more real root besides x = 5, and that root has multiplicity 2 To find it, we can use the quadratic formula with the coefficients and constant named according to the above scheme: x = [ b2 (b22 4a2c)1/2] / (2a2) Because the discriminant is equal to 0, we can simplify this to x = b2 / (2a2) Substituting in the values a2 = 9 and b2 = 24, we get x = ( 24) / [2 ( 9)] = 24/( 18) = 24/18 = 4/3 The roots of the cubic are therefore x = 5 or x = 4/3 The root x = 4/3 occurs with multiplicity 2 The solution set is X = {5, 4/3} 10 The new cubic, written in binomial-trinomial form, looks like this: (x + 3/2)(6x 2 4x + 2) = 0 Let s examine the discriminant d of the trinomial Setting a2 = 6, b2 = 4, and c = 2, we get d = b22 4a2c = ( 4)2 4 6 2 = 16 48 = 32 Because d < 0, the new cubic has only one real root, x = 3/2, exactly as the original cubic did (The two complex roots, however, differ in this equation compared with those in the final challenge in the chapter text For extra credit, you can verify this fact)
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1 In each of these situations, the trinomial can be factored into the square of a binomial Then that squared binomial is raised to the indicated power (a) Here is the original equation: (x 2 + 6x + 9)2 = 0
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In the trinomial, the coefficient of x 2 is 1, the coefficient of x is 6, and the standalone constant is 9 We must find a number, such that adding it to itself yields 6 while squaring it yields 9 That number is 3 The binomial is therefore (x + 3), and we have [(x + 3)2]2 = 0 which simplifies to (x + 3)4 = 0 (b) Here is the original equation: (x 2 4x + 4)3 = 0 In the trinomial, the coefficient of x 2 is 1, the coefficient of x is 4, and the constant is 4 We must find a number, such that adding it to itself yields 4 while squaring it yields 4 That number is 2 The binomial is therefore (x 2), and we have [(x 2)2]3 = 0 which simplifies to (x 2)6 = 0 (c) Here is the original equation: (16x 2 24x + 9)4 = 0 This trinomial is the square of (4x 3) Therefore, the original equation is equivalent to [(4x 3)2]4 = 0 which simplifies to (4x 3)8 = 0 2 In each case, we can remove the exponent from the binomial and then set it equal to 0, obtaining a first-degree equation The real root of the higher-degree equation is equal to the solution of the first-degree equation The multiplicity of the root is the value of the exponent n to which the binomial is raised
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694 Worked-Out Solutions to Exercises: s 21 to 29
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(a) The real root is found by solving x+3=0 That root is x = 3 Because the binomial is raised to the fourth power, this single real root has multiplicity 4 (b) The real root is found by solving x 2=0 That root is x = 2 Because the binomial is raised to the sixth power, this single real root has multiplicity 6 (c) The real root is found by solving 4x 3 = 0 We can add 3 to each side and then divide through by 4, obtaining the root x = 3/4 Because the binomial is raised to the eighth power, this single real root has multiplicity 8 3 In each of these situations, the trinomial can be factored into the product of two different binomials Then that product is raised to the indicated power (a) Here is the original equation: (x 2 3x + 2)2 = 0 In the trinomial, the coefficient of x 2 is 1, the coefficient of x is 3, and the standalone constant is 2 This trinomial factors into the product of (x 1) and (x 2) Therefore, the original equation can be rewritten as [(x 1)(x 2)]2 = 0 which can be further broken down to (x 1)2(x 2)2 = 0 (b) Here is the original equation: ( 3x 2 5x + 2)5 = 0 In the trinomial, the coefficient of x 2 is 3, the coefficient of x is 5, and the constant is 4 This trinomial factors into the product of (x + 2) and ( 3x + 1) Therefore, we can rewrite the original equation as [(x + 2)( 3x + 1)]5 = 0
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