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16 April, 11:57

How many real solutions does the equation 0.2x^5-2x^3+1.8x+k=0 have when k = 0?

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  1. 16 April, 12:16
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    When k = we have:-

    0.2x^5 - 2x^3 + 1.8x = 0

    so x = 0 is one solution

    taking 0.2x out we have:-

    0.2x (x^4 - 10x^2 + 9x) = 0

    factoring::-

    (x^2 - 9) (x^2 - 1) = 0

    this gives x = + / = 3 and x = + / - 1

    so there are 5 real solutions
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