# how to use barcode in c#.net Multiplying to confirm, we get (x + 5)(2x 2) = 0 2x 2x + 10x 10 = 0 2x 2 + 8x 10 = 0 in Software Print QR Code ISO/IEC18004 in Software Multiplying to confirm, we get (x + 5)(2x 2) = 0 2x 2x + 10x 10 = 0 2x 2 + 8x 10 = 0

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Multiplying to confirm, we get (x + 5)(2x 2) = 0 2x 2x + 10x 10 = 0 2x 2 + 8x 10 = 0
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The roots are found by solving these two first-degree equations: x+5=0 and 2x 2 = 0 The solution in the first case is easily seen to be x = 5, and in the second case the solution is x = 1 The roots of the quadratic are 5 and 1 The solution set is { 5, 1} 4 We want to morph the left side of the following quadratic into a product of binomials whose coefficients and constants in the binomials are all integers 12x 2 + 7x 10 = 0 In this case, the coefficient of x 2 is equal to 12 That means the general binomial factor form will look like one of these: (x + #)(12x + #) = 0 (2x + #)(6x + #) = 0 (4x + #)(3x + #) = 0 where # represents an integer (not necessarily the same one in each case) The product of these unknown integers is 10 As before, we can start plugging in integers whose product is 10, multiply the resulting product of binomials out on every attempt, and see what we get There are lots of choices here, and the process could take time Eventually we ll come up with (4x + 5)(3x 2) = 0 When we multiply this out, we find (4x + 5)(3x 2) = 0 12x + ( 8x) + 15x + ( 10) = 0 12x 2 + 7x 10 = 0
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To find the roots, we must solve 4x + 5 = 0
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and 3x 2 = 0 The solution to the first of these equations is derived like this: 4x + 5 = 0 4x = 5 x = 5/4 In the second case, the process is similar: 3x 2 = 0 3x = 2 x = 2/3 The roots of the quadratic are 5/4 and 2/3, and the solution set is { 5/4, 2/3} 5 We want to morph the following quadratic so the left side becomes a product of a binomial with itself: 16x 2 40x + 25 = 0 The coefficient of x 2 is equal to 16, and we know it has to be the square of the first term in the binomial That means the first term must be 4x or 4x The constant in the polynomial is equal to 25, and it must be the square of the constant in the binomial That means the constant in the binomial must be 5 or 5 The coefficient of x in the polynomial is negative, telling us that the coefficient and the constant in the binomial must have opposite signs As things work out, we get (4x 5)2 = 0 Checking to be sure this is right, we can multiply it out: (4x 5)(4x 5) = 0 16x + ( 20x) + ( 20x) + 25 = 0
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16x 2 40x + 25 = 0 It works! We can also use ( 4x + 5)2 = 0
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This equation is equivalent to the other one To show this, we can derive one squared binomial from the other: (4x 5)2 = (4x 5)(4x 5) = ( 1)2(4x 5)(4x 5) = ( 1)(4x 5)( 1)(4x 5) = ( 4x + 5)( 4x + 5) = ( 4x + 5)2 This duplicity occurs with all squared binomials It simply comes out of the fact that ( 1)2 = 1 6 To find the root of the quadratic, we can start with either of the binomial factor equations we found Let s use the first one: (4x 5)2 = 0 Taking the square root of both sides, we obtain 4x 5 = 0 We can add 5 to each side, getting 4x = 5 Dividing through by 4 gives us the root x = 5/4 The solution set is {5/4} Let s plug the root into the original quadratic to be sure that it works: 16x 2 40x + 25 = 0 16 (5/4)2 40 (5/4) + 25 = 0 16 25/16 50 + 25 = 0 25 50 + 25 = 0 25 + 25 = 0 0=0 7 We want to morph the following quadratic so we can get the left side into a product of a binomial with itself x 2 + 6x 7 = 0 We can add 16 to each side to get x 2 + 6x + 9 = 16 The left side can now be factored into a square of a binomial: (x + 3)2 = 16
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