Steps | Solution | ||
1. | Assume an expression of the form (ax+b)n, considering the power of x, that comes nearly as the expression. | In this case, let us assume (x+2)2 | |
2. | Compare them, to check the deficient terms. | In this case, we get: (x+2)2 = x2 + 4x + 4 And, the expression to solve: x2 + 3x + 2 Clearly, the deficient terms are: x + 2 |
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3. | Add the deficient terms to both side of the equation, to complete the assumed expression. | In this case, adding (x + 2): (x2 + 3x + 2) + (x + 2) = 0 + (x + 2) or, x2 + 4x + 4 = (x + 2) or, (x + 2)2 = (x + 2) |
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4. | Replace common terms with another variable, and solve. | In this case, let y = (x + 2). Then we get: y2 = y This solves, y = 0 and y = 1 |
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5. | Use solution to solve original variable. | In this case, y = (x + 2). y = 0, gives: x + 2 = 0; or, x = -2 And, y = 1, gives: x + 2 = 1; or, x = -1 Thus, we get: x = -2, x = -1, which is the answer! |
Let us assume (x+2)3: (x + 2)3 = x3 + 6x2 + 12x + 8 Clearly, the deficient terms are: (x + 2) Adding to both sides: (x3 + 6x2 + 11x + 6) + (x + 2) = 0 + (x + 2) or, (x + 2)3 = (x + 2) Let, y = (x + 2). Then, y3 = y solves for 0, 1 and -1 x + 2 = 0 gives: x = -2 Also, x + 2 = 1 gives: x = -1 And, x + 2 = -1 gives: x = -3 Thus, we get: x = -2, x = -1, x = -3 |
Let us assume (x+3)3: (x + 3)3 = x3 + 9x2 + 27x + 27 Clearly, the deficient terms are: (x2 + 10x + 17) Adding to both sides: (x3 + 8x2 + 17x + 10) + (x2 + 10x + 17) = 0 + (x2 + 10x + 17) or, (x + 3)3 = (x2 + 10x + 17) Let, y = (x + 3). Then, y3 = y2 + 4y – 4 This solves for 1, 2 and -2 x + 3 = 1 gives: x = -2 Also, x + 3 = 2 gives: x = -1 And, x + 3 = -2 gives: x = -5 Thus, we get: x = -2, x = -1, x = -5 |
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