# inequality distinct real roots

• Sep 7th 2009, 05:18 AM
mark
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
• Sep 7th 2009, 05:30 AM
mr fantastic
Quote:

Originally Posted by mark
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.
• Sep 7th 2009, 05:37 AM
mark
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