According to the picture, $x^2 - 4x + 4 = 32 + 4$ is written as $(x - 2)^2 = 36$. Can someone please show me the steps so I can learn to do this manually?
I don't understand how $x^2 - 4x + 4 = 36$ is rewritten as $(x - 2)^2 = 36$. What's the simplest way to do this?
The OP wrote in a comment: "I know that they are identical, but what is the working out to make $x^2−4x+4$ [equal] to $(x−2)^2$?"
It's four years later, but this is what I think she needed.
Use $FOIL$:
$$(x−2)^2 = (x-2)(x-2) = x^2 (First) -2x (Inside) - 2x (Outside) + 4 (Last)$$
$$ = x^2 - 4x + 4$$
For any $\;a\in\Bbb R\;$ :
$$x^2+ax=\left(x+\frac a2\right)^2-\frac{a^2}4$$
In your case, $\;a=4\;$ so
$$x^2-4x=(x-2)^2-4\implies 0= x^2-4x-32=(x-2)^2-4-32\implies (x-2)^2=36$$
$$x^2 - 4x + 4 = (x-2)^2$$
is an identity. Therefore, you can replace $x^2 - 4x + 4$ with $(x-2)^2$
For reference, the $-2$ in $x-2$ was chosen, because that is what you need to get a $-4x$ to appear as one of the terms when you expand $(x-2)^2$. The $+4$ in $x^2 - 4x + 4$ was chosen because that's what you need to get the identity above.
$$x^2-2ax-r=0\implies x^2-2\cdot x\cdot a+a^2=a^2+r$$
$$\implies(x-a)^2=a^2+r\implies x-a=\pm\sqrt{a^2+r}\implies x=\cdots$$
