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Topic: solve 3 (Read 279 times) |
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K Sengupta
Senior Riddler
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Re: solve 3
« Reply #1 on: Jul 2nd, 2007, 8:25am » |
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I assume that you want the real solutions to the given equation.
Letting a = V(2*x^2 -2x+12) and b = V(x^2 -5); we obtain:
(a-b)^3/(a(a^2+3*b^2)) = 2/9
Or, 7*a^3 – 27*a^2*b + 21a*b^2 – 9*b^3 = 0
Or, (a-3b)(7*a^2 – 6ab + 3*b^2) = 0
But, 7*a^2 – 6ab + 3*b^2 = 0 gives a/b equal to a complex quantity. But, by assumption, each of a and b is real. This is a contradiction.
Thus, a = 3b, giving:
2*x^2 – 2x + 12 = 9(x^2- 5)
Or, 7*x^2 + 2x – 57 = 0
Or, (x+3)(7x-19) = 0
Or, x = -3, 19 /7
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| « Last Edit: Jul 2nd, 2007, 8:35am by K Sengupta » |
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