# how to use barcode in c#.net Enter the Cubic in Software Creation QR Code ISO/IEC18004 in Software Enter the Cubic

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Let s solve a two-by-two system in which one equation is linear and the other is cubic Consider these: x 3 + 6x 2 + 14x y = 7 and 6x + 2y = 2
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First, we morph Again, it appears as if we ought to let x be the independent variable, and then derive two functions of that variable In the first equation, we can add 7 to each side, getting
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x 3 + 6x 2 + 14x y + 7 = 0 Then we can add y to each side and transpose the equation left-to-right, obtaining y as a function of x : y = x 3 + 6x 2 + 14x + 7 In the second equation, we can divide through by 2 to obtain 3x y = 1 Adding 1 to each side gives us 3x y + 1 = 0
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We can add y to each side and transpose the equation left-to-right, getting the function y = 3x + 1
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Next, we mix When we mix the independent-variable parts of the above functions, we obtain one equation in one variable:
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x 3 + 6x 2 + 14x + 7 = 3x + 1 If we subtract the quantity (3x + 1) from both sides, we get x 3 + 6x 2 + 11x + 6 = 0 This is a straightforward cubic equation in polynomial standard form The roots aren t obvious from casual inspection, but we can use the techniques from Chap 25 to solve it
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Next, we solve Now that we have derived a cubic equation in one variable, our mission is to find its roots We can use synthetic division several times to obtain factors Ultimately, we find that the cubic factors into
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(x + 1)(x + 2)(x + 3) = 0 The roots can be found by solving the three equations we get when we set each binomial equal to 0 Those roots are x = 1, x = 2, and x = 3 The y-values can be found by plugging these roots into either of the original functions Let s use the linear one; it s the less messy of the two! For x = 1, we have y = 3x + 1 = 3 ( 1) + 1 = 3 + 1 = 2 Now we know that our first solution is (x, y) = ( 1, 2) When x = 2, we have y = 3x + 1 = 3 ( 2) + 1 = 6 + 1 = 5
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458 More Two-by-Two Systems
Our second solution is (x, y) = ( 2, 5) Plugging in x = 3, we have y = 3x + 1 = 3 ( 3) + 1 = 9 + 1 = 8 Our third solution is (x, y) = ( 3, 8)
Finally, we check There are six arithmetic exercises to do! It s tedious, but if we want to be sure our solutions are right, it s mandatory We d better be careful with the signs, using negative additions rather than subtractions as much as possible! We check ( 1, 2) in the first original equation:
x 3 + 6x 2 + 14x y = 7 ( 1)3 + 6 ( 1)2 + 14 ( 1) ( 2) = 7 1 + 6 1 + ( 14) + 2 = 7 1 + 6 + ( 14) + 2 = 7 7 = 7 Next, we check ( 2, 5) in the first original equation: x 3 + 6x 2 + 14x y = 7 ( 2)3 + 6 ( 2)2 + 14 ( 2) ( 5) = 7 8 + 6 4 + ( 28) + 5 = 7 8 + 24 + ( 28) + 5 = 7 7 = 7 Next, we check ( 3, 8) in the first original equation: x 3 + 6x 2 + 14x y = 7 ( 3)3 + 6 ( 3)2 + 14 ( 3) ( 8) = 7 27 + 6 9 + ( 42) + 8 = 7 27 + 54 + ( 42) + 8 = 7 7 = 7 That completes the check for the original cubic Now we plug ( 1, 2) into the second original equation: 6x + 2y = 2 6 ( 1) + 2 ( 2) = 2 6 + ( 4) = 2 2=2