Author Topic: Desperate help (Read 563 times) Tweet Share

khalil · « **on:** September 07, 2009, 06:35:34 am »

$f(x)= (k+2)x^2+(6k-4)x+2$

If the turning points are (a,b) and b= $\frac{-(2-3k)^{2}+2(k+2)}{k+2}>0$

Find in terms of k {k: b>0}

I typed this b in the caclulator and got 0< k < 14/9 but the answers state this and k < -2. Where did they get k<-2 from?

cobby · « **Reply #1 on:** September 07, 2009, 07:29:41 am »

I typed that into my calculator and got :

$\frac {- \sqrt{97} - 13}{18} < k < \frac {\sqrt{97} - 13}{18} OR k < -2$

Did you type it in correctly?

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Author Topic: Desperate help (Read 563 times) Tweet Share

khalil

Desperate help

cobby

Re: Desperate help

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