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Math Help - nature of roots

  1. #1
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    nature of roots

    Determine the condition to be satisfied by k such that the expression 2x^2+6x+1+k(x^2+2) is positive for all x belongs to real numbers .

    Let 2x^2+6x+1+k(x^2+2)=0

    (2+k)x^2+6x+1+2k=0

    The graph must be a 'floating' graph on the positive side . I am not sure if i can say it that way. Thus , it has no real roots ..

    6^2-4(2+k)(1+2k)<0

    -8k^2-20k+24<0

    I got the wrong range of values for k . Where is my mistake ?
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  2. #2
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    Quote Originally Posted by thereddevils View Post
    Determine the condition to be satisfied by k such that the expression 2x^2+6x+1+k(x^2+2) is positive for all x belongs to real numbers .

    Let 2x^2+6x+1+k(x^2+2)=0

    (2+k)x^2+6x+1+2k=0
    First off, the leading coefficient must be positive in order for the expression to be positive for all x belonging to R, so
    2 + k > 0 => k > -2.

    The graph must be a 'floating' graph on the positive side . I am not sure if i can say it that way. Thus , it has no real roots ..

    6^2-4(2+k)(1+2k)<0

    -8k^2-20k+24<0

    I got the wrong range of values for k . Where is my mistake ?
    The last inequality should be
    -8k^2 - 20k + {\color{red}28} < 0

    So, divide both sides by -4:
    \begin{aligned}<br />
2k^2 + 5k - 7\;&{\color{red}>} 0 \\<br />
(2k + 7)(k - 1) &> 0<br />
\end{aligned}

    The solution set for k is
    (-\infty, -7/2) \cup (1, \infty).

    But I said earlier that k must be greater than -2, so intersect this with the solution set and you get
    (1, \infty)\;\text{or}\;k > 1.


    01
    Last edited by yeongil; July 18th 2009 at 07:26 AM.
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