f(x) = X[squared] -Kx + 9 , where K is a constant.

find the set values for K for which the equation f(x) = 0 has no real solutions.

and

given that k=4

express f(x) in the form (x - p)2 + q, where p and q are constants to be found.

thanks...

2. #1: Remember the discriminant?
Given the quadratic $\displaystyle ax^2 + bx + c = 0$, if $\displaystyle b^2 - 4ac \ {\color{red}<} \ 0$ then there are no real solutions to $\displaystyle x$.

So we must find when: $\displaystyle k^2 - 4(1)(9) < 0$

Can you solve this?

#2: $\displaystyle y = x^2 -kx + 9$ where $\displaystyle k=4$

We must complete the square. So we take the coefficient of $\displaystyle x$, divide it by 2, square it, and add+subtract it (remember, it's an equation so we can't randomly change one side):
$\displaystyle \begin{array}{rcl} y & = & \big(x^2 - kx + {\color{red}\left(\frac{k}{2}\right)^2}\big) + 9 - {\color{red}\left(\frac{k}{2}\right)^2} \\ & \vdots & \end{array}$

Can you finish?

so would it be:

1: -6<k<6

and

2: (x + 2)2 + 5

?

thanks