Solve the Quadratic Equation: x²-12x+36=0 for X

Let's solve the given equation:

$x^2-12x+36=0$

Note that we can factor the expression on the left side using the perfect square binomial formula:

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$

We'll do this using the fact that:

$36=6^2$Therefore, we'll represent the rightmost term as a squared term:

$x^2-12x+36=0 \\ \downarrow\\ \textcolor{red}{x}^2-24x+\textcolor{blue}{6}^2=0$

Now let's examine again the perfect square binomial formula mentioned earlier:

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$

And the expression on the left side of the equation that we got in the last step:

$\textcolor{red}{x}^2-\underline{12x}+\textcolor{blue}{6}^2=0$

Note that the terms $\textcolor{red}{x}^2,\hspace{6pt}\textcolor{blue}{6}^2$indeed match the form of the first and third terms in the perfect square binomial formula (which are highlighted in red and blue),

However, in order to factor this expression (on the left side of the equation) using the perfect square binomial formula mentioned, the remaining term must also match the formula, meaning the middle term in the expression (underlined):

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$

In other words - we'll ask if we can represent the expression on the left side of the equation as:

$\textcolor{red}{x}^2-\underline{12x}+\textcolor{blue}{6}^2=0\\ \updownarrow\text{?}\\ \textcolor{red}{x}^2-\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{6}}+\textcolor{blue}{12}^2=0$

And indeed it is true that:

$2\cdot x\cdot6=12x$

Therefore we can represent the expression on the left side of the equation as a perfect square binomial:

$\textcolor{red}{x}^2-\underline{2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{6}}+\textcolor{blue}{6}^2=0 \\ \downarrow\\ (\textcolor{red}{x}-\textcolor{blue}{6})^2=0$

From here we can take the square root of both sides of the equation (and don't forget that there are two possibilities - positive and negative when taking an even root of both sides of an equation), then we'll easily solve by isolating the variable:

$(x-6)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ x-6=\pm0\\ x-6=0\\ \boxed{x=6}$

Let's summarize the solution of the equation:

$x^2-12x+36=0 \\ \downarrow\\ \textcolor{red}{x}^2-2\cdot\textcolor{red}{x}\cdot\textcolor{blue}{6}+\textcolor{blue}{6}^2=0 \\ \downarrow\\ (\textcolor{red}{x}-\textcolor{blue}{6})^2=0 \\ \downarrow\\ x-6=0\\ \downarrow\\ \boxed{x=6}$

Therefore the correct answer is answer A.