The problem is completing the square of X^2 +6x +12=0.As I’m looking at each step they put add 9 on both sides. I’m puzzled on how did they get 9?
2 Answers
$\begingroup$You have $$x^2+6x=-12$$ and you want to make the left side equal to a square on the form $(ax+b)^2$, lets expand and compare $$(ax+b)^2=a^2x^2+2abx+b^2$$ we see that
- $a^2x^2=x^2$, so we may choose $a=1$ ($a=-1$ also works)
- $2bx=6x$, so we must have $b=3$
All of the above mean that we may write $$(x+3)^2=x^2+6x+9$$ but as you see we miss $9$, therefore we add it to complete the square.
$\endgroup$ $\begingroup$I'm not a fan of rote memorization but... this is how you do completing the square. You put all the terms to one side or another of the equation to get something like:
$x^2 + Kx = m$.
And then..... you ALWAYS .... divide the $K$ in half, square it and add it to both sides.
So $x^2 + 6x = -12$. Half of $6$ is $3$ and $3^2$ is $9$ so $x^2 + 6x + 9 = -12 + 9$ so $x^2 + 6x + 9 = (x+3)^2 = -3$ and $x + 3 =\pm \sqrt {-3}$.
Or if you had $x^2 + 14x = -13$... half of $14$ is $7$ and $7^2$ is $49$. So
$x^2 + 14x + 49 = -13 +49$ so $x^2 + 14x + 49 = (x+7)^2 = 36$ and $x +7 =\pm 6$.
Or if you have $x^2 + 5x = 41$ then half of $5$ is $\frac 52$ and $(\frac 52)^2 =\frac {25}4$ so $x^2 + 5x + \frac {25}4 = 41 + \frac {25}4$ so $x^2 + 5x + \frac {25}4 =(x +\frac 52)^2 = \frac {189}4$ so $x+\frac 52 =\pm \sqrt{\frac {189}4}$.
......
Now as to WHY why divide $K$ in half and square it.... keep reading.
=======
We have
- $x^2 + 6x = -12$
we want.....
- $(x + ???)^2 = !!!!!$
And
- $(x^2 + 6x + @@@@@@) = !!!!!$
where $(x + ???)^2 = x^2 + 6x + @@@@@@$
So how do we figure out what that $???$ ad $@@@@@$ are?
Well we have $(x + ???)^2 = (x^2 + 2??? + ???^2) = x^2 + 6x + @@@@@$.
So we must have $2??? = 6$. So that means $??? = 3$ and that means $@@@@@ = 3^2 = 9$.
.......
lt;dr
.......
$x^2 + 6x = -12$
so
$x^2 + 2\color{green}{(\frac 62)}x = -12$
$x^2 + 2\color{green}{(\frac 62)}x + (\color{green}{(\frac 62)})^2 =-12 + (\color{green}{(\frac 62)})^2$
$(x +\color{green}{(\frac 62)})^2 = -12 + (\color{green}{(\frac 62)})^2$
.....
ta;dl;l8r
$x^2 + 6x = -12$
$x^2 + 2\cdot\color{red}3 x = -12$
$x^2 + 2\cdot\color{red}3x + \color{red}3^2 = -12 +\color{red}3^2$
$(x + 3)^2 = -12 + 9 = -3$
$\endgroup$
