Sure! Let's solve the equation [tex]\( x^2 - 14x + 49 = 1 \)[/tex] by completing the square.
1. Start with the given equation:
[tex]\( x^2 - 14x + 49 = 1 \)[/tex]
2. First, move the constant term from the right side to the left side:
[tex]\( x^2 - 14x + 49 - 1 = 0 \)[/tex]
Simplifying this, we get:
[tex]\( x^2 - 14x + 48 = 0 \)[/tex]
3. Notice that we can rewrite the left-hand side as a complete square. To do this, we take the coefficient of [tex]\( x \)[/tex], which is -14, divide it by 2, and square the result:
[tex]\( \left(\frac{-14}{2}\right)^2 = 7^2 = 49 \)[/tex]
4. Now, rewrite the quadratic expression as a perfect square:
Since [tex]\( x^2 - 14x + 49 = (x - 7)^2 \)[/tex], we can rewrite the equation:
[tex]\( (x - 7)^2 = 1 \)[/tex]
5. Next, solve for [tex]\( x \)[/tex] by taking the square root of both sides:
[tex]\( x - 7 = \sqrt{1} \)[/tex] or [tex]\( x - 7 = -\sqrt{1} \)[/tex]
6. Solve each of these equations separately:
- For [tex]\( x - 7 = 1 \)[/tex]:
[tex]\( x = 1 + 7 \)[/tex]
[tex]\( x = 8 \)[/tex]
- For [tex]\( x - 7 = -1 \)[/tex]:
[tex]\( x = -1 + 7 \)[/tex]
[tex]\( x = 6 \)[/tex]
7. Therefore, the solutions to the equation [tex]\( x^2 - 14x + 49 = 1 \)[/tex] are:
[tex]\[
x = 8 \quad \text{and} \quad x = 6
\][/tex]
So, the solutions to the equation are:
[tex]\[ x = 8 \text{ and } x = 6 \][/tex]
