(x - 1)(x - 2)(x - 8)(x + 5) + 360 = 0
Solutions
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= -3, So we can write as |
[(x - 1)(x - 2)] [(x - 8)(x + 5)] + 360 = 0
(x2 - 2x - x + 2)(x2 + 5x - 8x - 40) + 360 = 0
(x2 - 3x + 2)(x2 - 3x - 40) + 360 = 0
(x2 - 3x + 2)(x2 - 3x + 2 - 42) + 360 = 0
Let x2 - 3x + 2 = t and Put in the equation
(t)(t - 42) + 360 = 0
t2 - 42t + 360 = 0
Using the quadratic formula, we have
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OR |
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t = 12 |
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t = 30 |
By Letting t = x2 - 3x + 2, Putting the value of t, we get
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x2 - 3x + 2 = 12 |
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x2 - 3x + 2 = 30 |
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x2 - 3x + 2 - 12 = 0 |
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x2 - 3x + 2 - 30 = 0 |
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x2 - 3x - 10 = 0 -------------------- (i) |
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x2 - 3x - 28 = 0 -------------------- (ii) |
By Equation (i) x2 - 3x - 10 = 0
Using the Quadratic formula, we have
By Equation (ii) x2 - 3x - 28 = 0
Using the Quadratic formula, we have
Hence the Solution Set of given equation is {-4, -2, 5, 7}