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Math Help - inequality distinct real roots

  1. #1
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    inequality distinct real roots

    hi, i've got 5x^2 - 7x + (2k - 3) = 0 and i've got it down to 109 - 40k > 0 if i move the -40k across i get 109 > 40k and the answer is \frac {109}{40} > k

    but if i move the 109 across and make it -40k > -109 i get k > \frac {109}{40} shouldn't it be the same whicever way i do it? what am i doing wrong?

    thanks
    Mark
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  2. #2
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    Quote Originally Posted by mark View Post
    hi, i've got 5x^2 - 7x + (2k - 3) = 0 and i've got it down to 109 - 40k > 0 if i move the -40k across i get 109 > 40k and the answer is \frac {109}{40} > k

    but if i move the 109 across and make it -40k > -109 i get k > \frac {109}{40} shouldn't it be the same whicever way i do it? what am i doing wrong?

    thanks
    Mark
    It would help if you posted the question rather than diving into what you've done and leaving us all to guess.

    I assume the question asks you to find the set of values of k for which 5x^2 - 7x + (2k - 3) = 0 has real solutions.

    -40k > -109 \Rightarrow 40k < 109 \Rightarrow k < \frac{109}{40} which is consistent with the first answer.

    Your mistake was not reversing the inequality sign when multiplying or dividing by a negative number.
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  3. #3
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    ahhh, i didn't know you had to do that. yeah sorry about lack of info. so its reversed if there's a negative number involved in division or multiplication. are there any exceptions? like double negatives? i suppose its unlikely that a double negative division would be an exception because that actually was the case in the question i just posted
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