# math

(please solve step by step)
solving equation (6x+2)/4 + (2x²-1)/(2x²+2) = (10x-1)/4x we got roots as

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1. (6x+2)/4 + (2x²-1)/(2x²+2) = (10x-1)/4x
Clear fractions by using the common denominator of 4x(x²+1). This requires that x≠0.

(6x+2)(x(x²+1)) + (2x²-1)(2x) = (10x-1)(x²+1)

(6x^4+2x^3+6x^2+2x) + (4x^3-2x) = (10x^3-x^2+10x-1)

6x^4 - 4x^3 + 7x^2 - 10x + 1 = 0 = 0

A little sniffing about with synthetic division shows that one root is x=1. So, now we have

(x-1)(6x^3+2x^2+9x-1) = 0

There are no other rational roots, so some kind of numeric method shows that another root is x = 0.1077

That leaves us with

6(x-1)(x-0.1077)(x^2+0.44x+1.55)

The quadratic has no real roots.

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posted by Steve

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