Not understanding the solution to how the square was completed

Question

$$f(x)=2x^2-12x+23$$

$$=2(x^{ 2 }-6x)+23$$

This part right here. Why is it $-2*9$? I got $2*9$ and it gave me completely different answer in the end. $$=2(x^{ 2 }-6x+9)+23-2*9$$

Answer 1

Here's the basic idea.

First move all constants to one side of the equation.

Then add or subtract a value from both sides of the equation such that a square is yielded on the side that contains the variables.

Here's one way of doing it $$ 2x^2-12x+23 =0 $$ $$ 2x^2-12x =-23 $$ $$ 2x^2-12x\color{red}{+18}=-23\color{red}{+18} $$ $$ 2(x^2-6x+9)=-5 $$ $$ 2(x-3)^2=-5 $$ $$ (x-3)^2=-\frac{5}{2} $$ Here's an other way of doing it $$ 2x^2-12x+23 =0 $$ $$ 2x^2-12x =-23 $$ $$ 2(x^2-6x)=-23 $$ $$ x^2-6x\color{red}{+9} =-\frac{23}{2}\color{red}{+9} $$ $$ (x-3)^2=-\frac{5}{2} $$ In general, $$ ax^2+bx+c=0, \mbox{for}\ a\ne 0 $$ Implies that $$ \left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2} $$ So in this case $$ a=2, b=-12, c=23 $$ Thus $$ \left(x-\frac{12}{2(2)}\right)^2=\frac{(-12)^2-4(2)(23)}{4(2)^2} $$ $$ \left(x-\frac{12}{4}\right)^2=\frac{144-8(23)}{4(4)} $$ $$ (x-3)^2=\frac{144-184}{16} $$ $$ (x-3)^2=-\frac{40}{16} $$ $$ (x-3)^2=-\frac{5}{2} $$ I hope this helps you understand.

Answer 2

Note that by adding the $9$ within the parenthesis, you have to subtract out $2\cdot 9$ in order to have an equivalent statement.

Answer 3

I would just like to say that when completing the square, all that you are really doing is writing things in a different form. You are not changing the original expression in any way

So if your original function is this

$f(x)=2x^2-12x+23$

Just remember that when completing the square, after every step it should be the same expression.

So as you said, the first step:

$2(x^2-6x)+23$

Note that if you expand it, you will get the original function.

The next step is to actually complete the square, and we do this by making sure that whatever is written in the bracket can also be written as a square.

This is the part you are having trouble with. We can write :

$2(x^2-6x+9)+23$

Here is the trick: spot the difference between this expression and the previous one. Yes, that's right. There is an extra +9. In maths you aren't allowed to just add something to an expression, so to compensate for this we have to somehow get rid of the +9. But note that the expression is $2(x^2-6x+9)$, we are multiplying it by 2. So that takes it from +9 to 2x9. So we now have a +18.

This part solves your problem. If you have a 'plus something', how do you cancel it out?

Simple, you 'minus something'.

$2(x-3)^2+23\color{red}{-18}$

The +18 cancels out with the -18 So you are left with:

$2(x-3)^2+5$

And you're done :)

Remember, if you want to check this, all you have to do is expand it - and see whether you have what you started with.

Answer 4

That's because you have added 9 in the parenthesis and the coefficient of the parenthesis is 2. So actually you have added 18 so you must subtract 18 as well. I hope it has helped you.