Answer :
Let's solve the given equation step by step to determine if Sarah's steps are accurate and to derive the final values of [tex]\( x \)[/tex].
Given equation:
[tex]\[ 2(x + 1)^2 + 6 = 134 \][/tex]
Step 1: Simplify the equation to isolate [tex]\((x + 1)^2\)[/tex]
[tex]\[ 2(x + 1)^2 + 6 = 134 \][/tex]
Subtract 6 from both sides:
[tex]\[ 2(x + 1)^2 = 128 \][/tex]
Divide both sides by 2:
[tex]\[ (x + 1)^2 = 64 \][/tex]
Step 2: Solve for [tex]\( x + 1 \)[/tex] by taking the square root of both sides
[tex]\[ x + 1 = \pm 8 \][/tex]
This step gives us two potential solutions because squaring both positive and negative roots will give the same result.
Step 3: Solve for [tex]\( x \)[/tex] in both cases
Case 1:
[tex]\[ x + 1 = 8 \][/tex]
Subtract 1 from both sides:
[tex]\[ x = 7 \][/tex]
Case 2:
[tex]\[ x + 1 = -8 \][/tex]
Subtract 1 from both sides:
[tex]\[ x = -9 \][/tex]
Step 4: State the solutions
[tex]\[ x = -9 \quad \text{or} \quad x = 7 \][/tex]
So, Sarah correctly identified the values for [tex]\( x \)[/tex] as [tex]\( -9 \)[/tex] and [tex]\( 7 \)[/tex]. Both solutions satisfy the original equation.
Given equation:
[tex]\[ 2(x + 1)^2 + 6 = 134 \][/tex]
Step 1: Simplify the equation to isolate [tex]\((x + 1)^2\)[/tex]
[tex]\[ 2(x + 1)^2 + 6 = 134 \][/tex]
Subtract 6 from both sides:
[tex]\[ 2(x + 1)^2 = 128 \][/tex]
Divide both sides by 2:
[tex]\[ (x + 1)^2 = 64 \][/tex]
Step 2: Solve for [tex]\( x + 1 \)[/tex] by taking the square root of both sides
[tex]\[ x + 1 = \pm 8 \][/tex]
This step gives us two potential solutions because squaring both positive and negative roots will give the same result.
Step 3: Solve for [tex]\( x \)[/tex] in both cases
Case 1:
[tex]\[ x + 1 = 8 \][/tex]
Subtract 1 from both sides:
[tex]\[ x = 7 \][/tex]
Case 2:
[tex]\[ x + 1 = -8 \][/tex]
Subtract 1 from both sides:
[tex]\[ x = -9 \][/tex]
Step 4: State the solutions
[tex]\[ x = -9 \quad \text{or} \quad x = 7 \][/tex]
So, Sarah correctly identified the values for [tex]\( x \)[/tex] as [tex]\( -9 \)[/tex] and [tex]\( 7 \)[/tex]. Both solutions satisfy the original equation.