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Math | Graded Assignment | Unit Test, Part 2 | Quadratic Equations
(Score for Question 2: [tex]$\qquad$[/tex] of 3 points)

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

[tex]\[
\begin{aligned}
x^2 - 6x - 7 &= 0 \\
x^2 - 6x &= 7 \\
x^2 - 6x + 9 &= 7 + 9 \\
(x - 3)^2 &= 16 \\
x - 3 &= 4 \\
x &= 7 \\
x - 3 &= -4 \\
x &= -1
\end{aligned}
\][/tex]

Is Josh's solution correct? Explain.

Answer:


Sagot :

Josh's solution for the equation [tex]\( x^2 - 6x - 7 = 0 \)[/tex] is incorrect. Let's examine why by solving the equation step-by-step and comparing it to Josh's solution:

Given the quadratic equation:
[tex]\[ x^2 - 6x - 7 = 0 \][/tex]

Step 1: Move the constant to the other side.
[tex]\[ x^2 - 6x = 7 \][/tex]

Step 2: Complete the square. To do this, we add and subtract [tex]\(\left( \frac{-6}{2} \right)^2 = 9\)[/tex] inside the equation:
[tex]\[ x^2 - 6x + 9 = 7 + 9 \][/tex]
[tex]\[ (x - 3)^2 = 16 \][/tex]

Step 3: Solve the resulting equation by taking the square root of both sides:
[tex]\[ x - 3 = \pm 4 \][/tex]

Step 4: Split into two separate equations:
[tex]\[ x - 3 = 4 \quad \text{or} \quad x - 3 = -4 \][/tex]

Step 5: Solve each equation for [tex]\(x\)[/tex]:
[tex]\[ x - 3 = 4 \quad \Rightarrow \quad x = 7 \][/tex]
[tex]\[ x - 3 = -4 \quad \Rightarrow \quad x = -1 \][/tex]

Therefore, the correct solutions to the equation [tex]\(x^2 - 6x - 7 = 0\)[/tex] are:
[tex]\[ x = 7 \quad \text{or} \quad x = -1 \][/tex]

Let's compare this with Josh's solution process:

Josh began correctly but made a critical mistake at the step where he solved [tex]\((x - 3)^2 = 16\)[/tex].

Instead of recognizing [tex]\(x - 3 = \pm 4\)[/tex], Josh incorrectly took each side:
[tex]\[ x - 3 = 16 \quad \text{and} \quad x - 3 = -16 \][/tex]

Which resulted in:
[tex]\[ x = 19 \quad \text{and} \quad x = -13 \][/tex]

These values are incorrect as solutions to the original equation. Therefore, Josh's solution is not correct. The correct solutions are [tex]\(x = 7\)[/tex] and [tex]\(x = -1\)[/tex].