1. anonymous

If $f(x)=12-\frac{ x ^{2} }{ 2 }$ and f(2k)=2k what is one possible value for k?

2. anonymous

solve 2k = 12 - (2k)^2/2

3. anonymous

y cant the answer be 5?

4. anonymous

probably because you solved it wrong.

5. anonymous
6. anonymous

|dw:1435945239538:dw|

7. anonymous

why cant it be that?

8. anonymous

when you divide by 2k you have 12/2k -2k^2/2k =12/2k - k =12/2k -k^2/k =12-2k^2/k

9. anonymous

okay ... go on

10. Owlcoffee

The problem with dividing an equations or even a function by the independant variable desired to solve for, you lose one of the values which is contained in the set of the solution. So, if you have: $f:f(x)=12- \frac{ x^2 }{ 2 }$ And we are asked to find the value that the function takes when x is equal to 2k, or in other words f(2k). $f(2k)=12-\frac{ (2k)^2 }{ 2 }$ $f(2k)=2k$ Then, knowing the value the function takes at the point "2k", we only need to find the value of "k" that satisfies this equation: $2k=12-\frac{ (2k)^2 }{ 2 }$ This has become a univariable equation but as a second degree polynomial expression, so what we will do is try to fix to the structure: $ax^2+bx+c=0$ in this case, "k" is the variable so we will treat it as if it was a popular known "x" first off, getting rid of the denominator by performing the common denominator operation: $2k=\frac{ 2(12)-(2k)^2 }{ 2 }$ And multiply both sides by "2": $2(2k)=2(12)-(2k)^2$ And now, simplifying: $4k=24-4k^2$ And taking it to the structure I mentioned earlier: $4k^2+4k-24=0$ And we can divide both sides by "4" in order to reduce the coefficients as much as possible: $\frac{ 4k^2+4k-24 }{ 4 }=\frac{ 0 }{ 4 }$ and simplifying: $k^2+k-6=0$ Now, in order to solve this we will have to use the general formula or also known as the "formula of bhaskara".

11. anonymous

so the way i did it was wrong?

12. Owlcoffee

Yes, unfortunately :(