Question 1095189: Can you please solve this equation using Cardano's method:
x^3+24x-25=0
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! The general form for a cubic equation is
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ax^3 + bx^2 + cx + d = 0
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the problem's cubic equation is
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x^3 + 24x - 25 = 0
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this is a depressed cubic equation, the general form is x^3 + Ax = B
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therefore we have
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x^3 + 24x = 25
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we determine s and t so that
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1) 3st = A
2) s^3 - t^3 = B
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Cardano's method tells us that x = s - t will be one of the roots of the cubic equation
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Solving the equation 1 for s and substituting into equation 2, we get
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(A/3t)^3 - t^3 = B
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simplifying this we get
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t^6 + Bt^3 - (A^3/27) = 0
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let u = t^3, then we have a quadratic equation
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u^2 + Bu - (A^3/27) = 0
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referring to our depressed cubic equation, we need s and t to satisfy
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3) 3st = 24
4) s^3 - t^3 = 25
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u^2 +25u -(24^3/27) = 0
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using quadratic formula, we have
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u = (-25 + square root(25^2 - 4 * 1 * (-512))) / (2*1) = 13.3505
u = (-25 - square root(25^2 - 4 * 1 * (-512))) / (2*1) = -38.3505
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we reject the negative solution for u, therefore
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t^3 = 13.3505
s^3 = t^3 + 25 = 13.3505 + 25 = 38.3508
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x = (38.3505)^(1/3) - (13.3505)^(1/3) = 1.000000955 approximately 1
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therefore one solution is x=1, the other two solutions are complex
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(x-1)(x^2+x+25) = 0 = x^3 + 24x - 25
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we see that the two solutions to x^2 + x + 25 = 0 (using quadratic formula) are
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x = (1/2)(-1-3isquare root(11)
x = (1/2)(-1+3isquare root(11)
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