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

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In the bottom equation, we can add the quantity (2 + j3) to each side, getting x = 2 + j3 The roots can be formally expressed this way: x = 2 j3 The solution set is X = {( 2 j3), (2 + j3)} 8 To get the polynomial form of the quadratic stated in Prob 7, we can multiply out the trinomial factors Here s the original equation again: (x + 2 + j3)(x 2 j3) = 0 Let s convert the subtractions in the second factor to negative additions individually to minimize the risk of getting the signs mixed up when we expand the equation into polynomial form That gives us (x + 2 + j3)[x + ( 2) + ( j3)] = 0 Now we can multiply out, obtaining x 2 + ( 2x) + ( j3x) + 2x + ( 4) + ( j6) + j3x + ( j6) + 9 =0 which simplifies to x 2 + ( j12) + 5 = 0 and further to x 2 + (5 j12) = 0 9 Here s the polynomial equation we derived It s interesting, because the coefficient of x is equal to 0, while the stand-alone constant is complex x 2 + (5 j12) = 0 Here are the roots we found: x = 2 j3 or x = 2 + j3 or x = 2 + j3
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680 Worked-Out Solutions to Exercises: s 21 to 29
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Plugging in the first root and converting the subtractions to additions, we can proceed like this, refining the equation step-by-step: [ 2 + ( j3)]2 + [(5 + ( j12)] = 0 [ 2 + ( j3)][ 2 + ( j3)] + [(5 + ( j12)] = 0 4 + j6 + j6 + ( j)29 + 5 + ( j12) = 0 Keeping in mind that ( j)2 = 1, we can simplify to 4 + j6 + j6 + ( 9) + 5 + ( j12) = 0 When we add all the terms on the left side, we get 0 = 0 The first root checks! Now we ll plug in the second root and convert the subtraction to negative addition We can proceed like this, step-by-step: (2 + j3)2 + [5 + ( j12)] = 0 (2 + j3)(2 + j3) + [5 + ( j12)] = 0 4 + j6 + j6 + j 29 + 5 + ( j12) = 0 4 + j6 + j6 + ( 9) + 5 + ( j12) = 0 That s the same equation we got when we plugged in the other root When we add all the terms on the left side, we get 0 = 0 The second root checks! 10 We ve been told to convert the following equation, which consists of two trinomial factors, into the polynomial standard form for a quadratic: (j2x + 2 + j3)( j5x + 4 j5) = 0 This might look daunting at first, but our assigned task is merely a matter of multiplying things out, taking care with the signs, and not rushing it! Let s begin by converting the subtraction in the second factor to addition That gives us (j2x + 2 + j3)[( j5x + 4 + ( j5)] = 0 We can use the product of sums rule, remembering that j j = 1 It goes like this: (j2x)( j5x) + j8x + (j2x)( j5) + ( j10x) + 8 + ( j10) + (j3)( j5x) + j12 + (j3)( j5) =0 Simplifying the individual terms, we get 10x 2 + j8x + 10x + ( j10x) + 8 + ( j10) + 15x + j12 + 15 =0
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