anonymous
  • anonymous
Please help =*(
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
If \[f(x)=12-\frac{ x ^{2} }{ 2 }\] and f(2k)=2k what is one possible value for k?
anonymous
  • anonymous
solve 2k = 12 - (2k)^2/2
anonymous
  • anonymous
y cant the answer be 5?

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anonymous
  • anonymous
probably because you solved it wrong.
anonymous
  • anonymous
k = -3 or 2 https://www.wolframalpha.com/input/?i=2k+%3D+12+-+%282k%29%5E2%2F2
anonymous
  • anonymous
|dw:1435945239538:dw|
anonymous
  • anonymous
why cant it be that?
anonymous
  • 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
anonymous
  • anonymous
okay ... go on
Owlcoffee
  • 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".
anonymous
  • anonymous
so the way i did it was wrong?
Owlcoffee
  • Owlcoffee
Yes, unfortunately :(

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