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].
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].
