SOLUTION: I had a question in my exam today and i couldnt answer it fully can you please help Qeustion)a) find the set of values of 'k' for which the equation 'f(x)=0' has no real sol

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Question 237368: I had a question in my exam today and i couldnt answer it fully
can you please help
Qeustion)a)
find the set of values of 'k' for which the equation 'f(x)=0' has no real solutions.
answer)
f(x)=x^2-kx+9, where k is a constant
b^2-4ac<0
k^2-4*1*9=k^2-36<0
k^2<36
k<6
-6 (-6,6)
that was fairly simple, however i struggled on the next part
Question)b)
Given that k=4,
express f(x) in the form (x-p)^2+q, where p and q are constants to be found.
answer)
Please help with this question because i really was angry when i couldnt complete it in my exam
thank you for your cooporation.
by the way i used '^' to show power of

Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website!
b)

Start with the given function.

Plug in

Now let's complete the square.

To do so, take half of the coefficient to get . In other words, .

Now square to get . In other words,

Now add and subtract . Make sure to place this after the "x" term. Notice how . So the expression is not changed.

Group the first three terms.

Factor to get .

Combine like terms.

So after completing the square, transforms to . So .

So is equivalent to .

Now is in the form where p=2 and q=5.