Convert the L.H.S to the form of the R.H.S by completing the square. Given that $x^2-3x+5=(x-p)^2+q$ for all values of $x$, calculate the value of $p$ and of $q$.
A book example tells me to firstly convert the L.H.S to the form of the R.H.S by completing the square.
How to convert it? Can anyone explain me how to do it? Can you show me the steps? HELP!
 A: $x^2−3x+5=(x-1.5)^2-2.25+5=(x-1.5)^2+2.75$
Motivation: So look at the expansion of $(x+a)^2$, this is $x^2+2ax+a^2$, can you see the reason why we half the coefficient of the x and why I subtracted by $2.25$?
A: You always have to take half of the linear coefficient, put that in the bracket and subtract the square of half the coefficient to make up for it:
$$\begin{align}
x^2\color{red}{-3}x+5 &=\left(x+\frac{\color{red}{-3}}{2}\right)^2-\left(\frac{\color{red}{-3}}{2}\right)^2+5\\
&=\left(x-\frac{3}{2}\right)^2 - \frac{9}{4} + \frac{20}{4}\\
&=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}
\end{align}
$$
The reason for this is the binomial formula:
$$(x+\color{red}{a})^2=x^2+2\color{red}{a}x+\color{red}{a}^2$$
A: $x^2-3x+5$
$=x^2-3x+\frac94-\frac94+5$
$=(x-\frac32)^2+\frac{11}4$
A: You must first make sure the LHS is in standard form and is a monomial (the coefficient of the leading term must be 1) then you would complete the square by taking the "b" term and dividing it by 2 and squaring it and adding it to the equation but also subtracting it (basically you're adding 0). You should have something of the form $(ax^2+bx+(b/2)^2)-(b/2)^2$. From this you can see the values for p and q once you realize you can factor the polynomial in parenthesis. 
