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    The equation x^"+5kx+2k=0, where k is a constant, has real roots.

    (a) Prove that k(25k-8) is greater than or equal to 0. (done this, see below)

    (b) Hence find the set of possible values of k

    (c) Write down the values of k for which the equation x^2+5kx+2k=0 has EQUAL roots.

    Part a-

    b^2 is greater than 4ac
    therefore 25k^2 is greater than or equal to 8k
    therefore, once factorised, you get k(25k-8) is greater than or equal to zero.
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    You have k(25k-8)>=0, so find the values of k that make this inequality true. For part (c) you want b^2 = 4ac.
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    Well can't do part (b). Any hints? I have k is greater than 8/25, not sure if this is correct?- and don't know how to get the other limiting value.
    Thanks
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    (Original post by Lawbutwhere?)
    Well can't do part (b). Any hints? I have k is greater than 8/25, not sure if this is correct?- and don't know how to get the other limiting value.
    Thanks
    k>=8/25 or k=<0
 
 
 
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Updated: April 10, 2006
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