College

This is Josh's solution for the equation [tex]x^2 - 6x - 7 = 0[/tex]:

[tex]
\begin{aligned}
x^2 - 6x - 7 & = 0 \\
x^2 - 6x & = 7 \\
x^2 - 6x + 9 & = 7 + 9 \\
(x - 3)^2 & = 16 \\
x - 3 & = 16 \\
x & = 19 \\
x - 3 & = -16 \\
x & = -13
\end{aligned}
[/tex]

Is Josh's solution correct? Explain.

Answer :

Josh's solution isn't correct. Let's go through the correct steps to solve the quadratic equation [tex]\(x^2 - 6x - 7 = 0\)[/tex] using the method of completing the square:

1. Start with the original equation:

[tex]\[
x^2 - 6x - 7 = 0
\][/tex]

2. Move the constant term to the other side to prepare for completing the square:

[tex]\[
x^2 - 6x = 7
\][/tex]

3. Complete the square on the left side:

- Take half of the coefficient of [tex]\(x\)[/tex], which is [tex]\(-6\)[/tex], to get [tex]\(-3\)[/tex].
- Square [tex]\(-3\)[/tex] to get [tex]\(9\)[/tex].
- Add [tex]\(9\)[/tex] to both sides of the equation to maintain the balance:

[tex]\[
x^2 - 6x + 9 = 7 + 9
\][/tex]

- This gives:

[tex]\[
(x-3)^2 = 16
\][/tex]

4. Solve the equation by taking the square root of both sides:

- The equation [tex]\((x-3)^2 = 16\)[/tex] implies:

[tex]\[
x - 3 = \pm 4
\][/tex]

- Solve for [tex]\(x\)[/tex] in both cases:

- [tex]\(x - 3 = 4 \Rightarrow x = 7\)[/tex]
- [tex]\(x - 3 = -4 \Rightarrow x = -1\)[/tex]

Therefore, the correct solutions to the equation [tex]\(x^2 - 6x - 7 = 0\)[/tex] are [tex]\(x = 7\)[/tex] and [tex]\(x = -1\)[/tex].

In Josh's solution, he made an error by incorrectly solving the equation [tex]\((x-3)^2 = 16\)[/tex] as [tex]\(x - 3 = 16\)[/tex] and [tex]\(x - 3 = -16\)[/tex], which led him to find the incorrect solutions [tex]\(x = 19\)[/tex] and [tex]\(x = -13\)[/tex].