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

Q: How do I know if a quadratic is a perfect square?

Check if the first and last terms are perfect squares, and the middle term equals $ 2ab $. For $ x^2 - 12x + 36 $: $ x^2 $ and $ 36 = 6^2 $ are squares, and $ 12x = 2 \cdot x \cdot 6 $!

Q: What if the middle term is positive instead of negative?

Then you have $ (a + b)^2 = a^2 + 2ab + b^2 $ instead! For example, $ x^2 + 12x + 36 = (x + 6)^2 $. The sign in the factored form matches the middle term's sign.

Q: Why does (x-6)² = 0 have only one solution?

When you take the square root: $ x - 6 = \pm 0 $. But $ +0 = -0 = 0 $, so both give $ x = 6 $. This is called a repeated root or double root.

Q: Can I still use the quadratic formula on perfect squares?

Yes, but it's unnecessarily complicated! The quadratic formula will give $ x = \frac{12 \pm 0}{2} = 6 $, but factoring is much faster and less error-prone.

Q: How do I check my answer is correct?

Substitute $ x = 6 $ back: $ 6^2 - 12(6) + 36 = 36 - 72 + 36 = 0 $ ✓. The left side equals zero, confirming our solution!

Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:06 Let's find X. Ready?

00:09 First, we'll use shortcut multiplication and watch the coefficients. Got it?

00:31 We need two numbers.

00:37 Their sum should be B, and their product should be C.

00:42 These are our numbers. Great job!

00:46 So, we'll place these numbers in parentheses.

00:53 Next, find solutions that make each factor zero.

00:58 Since they are equal, the solution is the same.

01:03 Let's isolate X.

01:06 And that's how we solve this problem. Well done!

Step-by-step written solution

Follow each step carefully to understand the complete solution

1

Understand the problem

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

Determine the value of X:

2

Step-by-step solution

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 obtained 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.

3

Final Answer

6

Key Points to Remember

Essential concepts to master this topic

Perfect Square Pattern: Recognize $a^2 - 2ab + b^2 = (a-b)^2$
Factoring Technique: $x^2 - 12x + 36 = (x-6)^2$ since $2 \cdot x \cdot 6 = 12x$
Verification Check: Square root both sides: $(x-6)^2 = 0$ gives $x = 6$ ✓

Common Mistakes

Avoid these frequent errors

Using quadratic formula instead of recognizing perfect square pattern
Don't automatically jump to $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ = complex calculations! This wastes time and increases error chances. Always check if the quadratic is a perfect square trinomial first - it makes solving much simpler.

Practice Quiz

Test your knowledge with interactive questions

$$ x^2+6x+9=0 $$

What is the value of X?

FAQ

Everything you need to know about this question

How do I know if a quadratic is a perfect square?

+

Check if the first and last terms are perfect squares, and the middle term equals $2ab$ . For $x^2 - 12x + 36$ : $x^2$ and $36 = 6^2$ are squares, and $12x = 2 \cdot x \cdot 6$ !

What if the middle term is positive instead of negative?

+

Then you have $(a + b)^2 = a^2 + 2ab + b^2$ instead! For example, $x^2 + 12x + 36 = (x + 6)^2$ . The sign in the factored form matches the middle term's sign.

Why does (x-6)² = 0 have only one solution?

+

When you take the square root: $x - 6 = \pm 0$ . But $+0 = -0 = 0$ , so both give $x = 6$ . This is called a repeated root or double root.

Can I still use the quadratic formula on perfect squares?

+

Yes, but it's unnecessarily complicated! The quadratic formula will give $x = \frac{12 \pm 0}{2} = 6$ , but factoring is much faster and less error-prone.

How do I check my answer is correct?

+

Substitute $x = 6$ back: $6^2 - 12(6) + 36 = 36 - 72 + 36 = 0$ ✓. The left side equals zero, confirming our solution!

🌟 Unlock Your Math Potential

Get unlimited access to all 18 Factorization questions, detailed video solutions, and personalized progress tracking.

📹

Unlimited Video Solutions

Step-by-step explanations for every problem

📊

Progress Analytics

Track your mastery across all topics

🚫

Ad-Free Learning

Focus on math without distractions

No credit card required • Cancel anytime

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

Quadratic Equations with Perfect Square Factoring

❤️ Continue Your Math Journey!