How do you solve $f\left( x \right) = 2{x^2} - 12x + 17$ by completing the square?

Hint: In this question, we are given an equation and we have been asked to solve the equation by using the method of completing the square. In this method, first we have to ensure that ${x^2}$ has no coefficient. If it has a coefficient, then we multiply the entire equation by that coefficient. Then, halve the coefficient of $x$, square it and, add and subtract it in the equation. Then, form the square. Keep the equation equal to 0 and find the value of $x$.

Complete step-by-step solution:
We are given an equation $f\left( x \right) = 2{x^2} - 12x + 17$ and we have to solve it using completing the square.
This method is a little complicated as compared to other methods.
So, I have mentioned the steps along with doing the calculation.
Step 1: Divide the entire equation by the coefficient of ${x^2}$ (This step is to be followed if ${x^2}$ has a coefficient.)
$ \Rightarrow f\left( x \right) = 2{x^2} - 12x + 17$
Dividing the entire equation by$2$
$ \Rightarrow f\left( x \right) = \dfrac{{2{x^2}}}{2} - \dfrac{{12x}}{2} + \dfrac{{17}}{2}$
On simplification we get,
$ \Rightarrow f\left( x \right) = {x^2} - 6x + \dfrac{{17}}{2}$
Step 2: Halve the coefficient of$x$.
Square the resultant and add and subtract it to the equation.
$ \Rightarrow f\left( x \right) = {x^2} - 6x + \dfrac{{17}}{2} + {\left( 3 \right)^2} - {\left( 3 \right)^2}$
Step 3: Complete the square by looking for the terms like ${a^2}$, ${b^2}$ and $2ab$.
If we observe, ${a^2} = {x^2}$, ${b^2} = {3^2}$ and $2ab = - 6x\left( { = - 2 \times 3 \times x} \right)$
Also, it must be noted that since the value of our $2ab$ is negative, our square will be of negative term, that is, ${(a - b)^2}$.
$ \Rightarrow f\left( x \right) = {\left( {x - 3} \right)^2} + \dfrac{{17}}{2} - 9$
Simplifying it further,
$ \Rightarrow f\left( x \right) = {\left( {x - 3} \right)^2} + \dfrac{{17}}{2} - \dfrac{{18}}{2}$
Let us subtract the term and we get
$ \Rightarrow f\left( x \right) = {\left( {x - 3} \right)^2} - \dfrac{1}{2}$
Step 4: Find the value of $x$.
Now, we will keep our equation equal to “$0$”.
$ \Rightarrow {\left( {x - 3} \right)^2} - \dfrac{1}{2} = 0$
Shifting the terms to find the value of$x$,
$ \Rightarrow {\left( {x - 3} \right)^2} = \dfrac{1}{2}$
Square rooting both the sides,
$ \Rightarrow \sqrt {{{\left( {x - 3} \right)}^2}} = \sqrt {\dfrac{1}{2}} $
We get,
$ \Rightarrow x - 3 = \pm \dfrac{1}{{\sqrt 2 }}$
On add 3 on both sides we get
$ \Rightarrow x = \pm \dfrac{1}{{\sqrt 2 }} + 3$

Hence, $x = 3 \pm \dfrac{1}{{\sqrt 2 }}$ is our final answer

Note: How will we know that the square which will be formed is ${\left( {a + b} \right)^2}$ or ${\left( {a - b} \right)^2}$ ?
Identify the term which will be ours $2ab$. Then see whether it is negative or positive.
If it is positive, then the square formed will be of ${\left( {a + b} \right)^2}$
If the term is negative, then the square formed is of ${\left( {a - b} \right)^2}$.
This is a very important rule and should be understood as writing the wrong square will not give you any marks.