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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: khalil on September 07, 2009, 06:35:34 am

Title: Desperate help
Post by: 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?
Title: Re: Desperate help
Post by: cobby 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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