Sure, let's solve the given equation step-by-step and determine whether the solution is extraneous or not.
Given the equation:
[tex]\[ \sqrt{2x + 1} = 3 \][/tex]
Step 1: Square both sides of the equation to eliminate the square root.
[tex]\[ (\sqrt{2x + 1})^2 = 3^2 \][/tex]
[tex]\[ 2x + 1 = 9 \][/tex]
Step 2: Solve for [tex]\( x \)[/tex].
To isolate [tex]\( x \)[/tex], subtract 1 from both sides:
[tex]\[ 2x + 1 - 1 = 9 - 1 \][/tex]
[tex]\[ 2x = 8 \][/tex]
Then, divide both sides by 2:
[tex]\[ \frac{2x}{2} = \frac{8}{2} \][/tex]
[tex]\[ x = 4 \][/tex]
Step 3: Substitute [tex]\( x = 4 \)[/tex] back into the original equation to check if it holds true.
Original equation:
[tex]\[ \sqrt{2x + 1} = 3 \][/tex]
Substitute [tex]\( x = 4 \)[/tex]:
[tex]\[ \sqrt{2(4) + 1} = \sqrt{8 + 1} \][/tex]
[tex]\[ \sqrt{9} = 3 \][/tex]
[tex]\[ 3 = 3 \][/tex]
Since both sides of the original equation are equal when [tex]\( x = 4 \)[/tex], the solution [tex]\( x = 4 \)[/tex] is correct and not extraneous.
Final Answer:
[tex]\[ x = 4 \][/tex]
Solution is not extraneous.
