SOLUTION: Demonstrate that the roots of the following equation are rational abc�x� + 3a�cx + b�cx - 6a�

SOLUTION: Demonstrate that the roots of the following equation are rational abc�x� + 3a�cx + b�cx - 6a� - ab + 2b� = 0

Question 1117462: Demonstrate that the roots of the following equation are rational
abc�x� + 3a�cx + b�cx - 6a� - ab + 2b� = 0

Found 2 solutions by Edwin McCravy, greenestamps:
Answer by Edwin McCravy(20055) (Show Source): You can put this solution on YOUR website!

abc�x� + 3a�cx + b�cx - 6a� - ab + 2b� = 0 We put the left side abc�x� + 3a�cx + b�cx - 6a� - ab + 2b� in quadratic form: Ax� + Bx + C (abc�)x� + (3a�cx + b�x) - 6a�-ab+2b� (abc�)x� + (3a�c+b�)x + (-6a�-ab+2b�) The x� term is abc�x� which has most likely factorization (acx)�(bcx) So I'll try to factor it like this: [acx ][bcx ] The constant term is -6a� - ab + 2b� That factors, take out a - sign -(6a� + ab - 2b�) and again as -(2a - b)(3a + 2b) So I'll try to factor the original left side as one of these possible factorizations: (1). [acx + (2a-b)][bcx - (3a+2b)] (2). [acx - (2a-b)][bcx + (3a+2b)] (3). [acx + (3a+2b)][bcx - (2a-b)] (4). [acx - (3a+2b)][bck + (2a-b)] I could multiply those out, but it would be easier to choose arbitrary easy values for the letters and check their results with the results of the original left side. Just choose one of the letters as 0 and the rest as 1, so hopefully only one of the four possible factorizations will be the same as when we substitute them in the original left side. If two or more are the same as that value, we'll have to make up different values for the letters. a = 0, b = c = x = 1 Substituting in possible factorization (1). [acx + (2a-b)][bcx - (3a+2b)] [0�1�1 + (2�0-1)][1�1�1 - (3�0+2�1)] = [0-1][1-2] = [-1][-1] = 1 Substituting in possible factorization (2). [acx - (2a-b)][bcx + (3a+2b)] [0�1�1 - (2�0-1)][1�1�1 + (3�0+2�1)] = [0+1][1+2] = [1][3] = 3 Substituting in possible factorization (3). [acx + (3a+2b)][bcx - (2a-b)] [0�1�1 + (3�0+2�1)][1�1�1 - (2�0-1)] = [0+2][1+1] = [2][2] = 4 Substituting in possible factorization (4). [acx - (3a+2b)][bcx + (2a-b)] [0�1�1 - (3�0+2�1)][1�1�1 + (2�0-1)] = [0-2][1-1] = [-2][0] = 0 That's good. They are all different. Now we can tell which one is probably the correct factorization by substituting those same values in the original left side: a = 0, b = c = x = 1 abc�x� + 3a�cx + b�cx - 6a� - ab + 2b� 0�1�1��1� + 3�0��1�1 + 1��1�x - 6�0� - 0�1 + 2�1� 0 + 0 + 1 - 0 - 0 + 2 = 3 So only possible factorization (2) can be the correct one. But we really haven't shown that it is the correct one, but only if any are correct, it's possible factorization (2). So let's be sure: (2). [acx - (2a-b)][bcx + (3a+2b)] So pick arbitrary values for the letters, say a=3, b=4, c=2, x=1 Substituting in (2) (2). [acx - (2a-b)][bcx + (3a+2b)] [3�2�1 - (2�3-4)][4�2�1 + (3�3+2�4)] [6 - (6-4)][8 + (9+8)] [6 - 2][8 + 17] [4][25] 100 Substituting in the original left side: a=3, b=4, c=2, x=1 3�4�2��1� + 3�3��2�1 + 4��2�1 - 6�3� - 3�4 + 2�4� 3�4�4�1 + 3�9�2�1 + 16�2�1 - 6�9 - 3�4 + 2�16 48 + 54 + 32 - 54 - 12 + 32 100 Since we got the same 100, we are 99.999% sure that (2) is the correct factorization. We can simplify (2) slightly as (2). [acx - (2a-b)][bcx + (3a+2b)] [acx - 2a + b][bcx + 3a + 2b] So the original equation: abc�x� + 3a�cx + b�cx - 6a� - ab + 2b� = 0 becomes [acx - 2a + b][bcx + 3a + 2b] = 0 Using the zero factor property: acx - 2a + b = 0; bcx + 3a + 2b = 0 acx = 2a - b x = <-- one root bcx + 3a + 2b = 0 bcx = -3a - 2b x = <-- other root If a,b and c are rational and not 0, then both roots are rational also. Edwin

Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

Part of my solution to the problem is basically the same as the other tutor's; the other part is far different.

As he says, put the equation in pure quadratic form:

In a quadratic equation

the sum of the roots is and the product of the roots is .

In this equation, the product of the roots is

or

With abc^2 as the denominator in the product of the two roots, the denominators of the two roots are most likely ac and bc. There are then four possible pairs of roots:

(1) and
(2) and
(3) and
(4) and

To find which is the correct pair of roots, we use the fact that the sum of the roots has to be

The second pair of roots gives us this sum:

So the two roots are

and

And as long as none of a, b, or c is 0, those roots are rational.
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