How to factorize $2x^2-9x+9$ by completing the square? I know that $x^2-bx+c=(x-k)^2=x^2-2kx+k^2$ if it is a complete square. If not we create one by adding and subtracting $\left(\frac{b}{2}\right)^2$
I tried $$2\left(x^2-\frac{9}{2}x+\frac{9}{2}\right)=2\left(x^2-\frac{9}{2}x+\left(\frac{9/2}{2}\right)^2-\left(\frac{9/2}{2}\right)^2+\frac{9}{2}\right)$$
What am I missing and what is the best way to factorize when we have coefficients?(Is this the right term, ($a x^2$)?)
 A: Notice that $$x^2-\frac{9}{2}x+\left(\frac{9}{4}\right)^2-\left(\frac{9}{4}\right)^2+\frac{9}{2}=\left(x-\frac{9}{4}\right)^2-\left(\frac{3}{4}\right)^2=\left(x-\frac{9}{4}-\frac{3}{4}\right)\left(x-\frac{9}{4}+\frac{3}{4}\right)$$
A: $$
\begin{align*}
2x^2 - 9x + 9 &= 2\left(x^2 - \frac{9}{2}x + \frac{9}{2}\right)
\\&= 2\left(x^2 - 2\frac{9}{4}x + \frac{9}{2}\right)
\\&= 2\left(x^2 - 2\frac{9}{4}x + \frac{9}{2} + \left(\frac{9}{4}\right)^2 - \left(\frac{9}{4}\right)^2\right)
\\&= 2\left(\left(x-\frac{9}{4}\right)^2 + \frac{9}{2} - \left(\frac{9}{4}\right)^2\right)
\\&= 2\left(\left(x-\frac{9}{4}\right)^2 - \left(-\frac{9}{2} + \left(\frac{9}{4}\right)^2\right)\right)
\\&= 2\left(\left(x-\frac{9}{4}\right)^2 - \sqrt{-\frac{9}{2} + \left(\frac{9}{4}\right)^2}^2\right)
\\&= 2\left(\left(x-\frac{9}{4}\right) - \sqrt{-\frac{9}{2} + \left(\frac{9}{4}\right)^2}\right)\left(\left(x-\frac{9}{4}\right) + \sqrt{-\frac{9}{2} + \left(\frac{9}{4}\right)^2}\right)
\\&= 2\left(x-\frac{9}{4} - \sqrt{-\frac{9}{2} + \left(\frac{9}{4}\right)^2}\right)\left(x-\frac{9}{4} + \sqrt{-\frac{9}{2} + \left(\frac{9}{4}\right)^2}\right)
\end{align*}
$$
A: $$2x^2-9x+9\tag1$$
Factorise 
$$2\left(x^2-{9\over 2}x+{9\over 2}\right)\tag2$$
$$2\left(x^2-{9\over 2}x+{9\over 2}\right)=2\left[\left(x-{9\over 4}\right)^2+{9\over 2}-\left(9\over 4\right)^2\right]\tag3$$
$$2\left(x^2-{9\over 2}x+{9\over 2}\right)=2\left[\left(x-{9\over 4}\right)^2-{9\over 16}\right]\tag4$$
$$2x^2-9x+9=2\left(x-{9\over 4}\right)^2-{9\over 8}\tag5$$
A: What you have done is correct so far. Nice job! Next time, you could divide both sides of $2x^2-9x+9 = 0$ to get $x^2-\frac{9}{2}x + \frac{9}{2}=0$.
From your method, we have:
$$(x-\frac{9}{4})^2 - (\frac{9}{4})^2+\frac{9}{2} =0$$
$$(x-\frac{9}{4})^2 = \frac{81}{16} - \frac{9}{2}$$
$$(x - \frac{9}{4})^2 = \frac{9}{16}.$$
The next step is very important. When taking the square root, there are always two solutions — one positive and one negative.
$$x - \frac{9}{4} = \color{red}{±} \color{black}{\frac{3}{4}}$$
From here you can find the two values of $x$, and then factorise.
A: HINT
We have 
$$2\left(x^2-\frac{9}{2}x+\frac{9}{2}\right)=2\left(x-\frac{9}{4}\right)^2-\frac{81}{8}+9=2\left(x-\frac{9}{4}\right)^2-\frac98$$
A: Make the polynomial equal to $0$ to find the roots so that you can factorise.
$$2x^2-9x+9=0$$
$$x^2-\frac{9x}{2}+\frac{9}{2}=0$$
$$x^2-\frac{9x}{2}=\frac{-9}{2}$$
To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of b.
$$(\frac{b}{2})^2=(\frac{-9}{4})^2$$
Add the term to each side of the equation.
You get:
$$(x-\frac{9}{4})^2=\frac{9}{16}$$
$$x=\frac{9}{4} \pm \frac{3}{4}$$
$$\implies x=3,\frac{3}{2}$$
Hence, ($x$-$\frac{3}{2}$)($x$-$3$) is your answer.
