To determine whether the student's conclusion that [tex]\( x = 4 \)[/tex] is a solution to the equation [tex]\(\sqrt{2x + 1} + 3 = 0\)[/tex], let's follow these steps:
1. Substitute [tex]\( x = 4 \)[/tex] into the equation:
[tex]\[
\sqrt{2(4) + 1} + 3 = 0
\][/tex]
2. Simplify inside the square root:
[tex]\[
\sqrt{2 \cdot 4 + 1} + 3 = \sqrt{8 + 1} + 3 = \sqrt{9} + 3
\][/tex]
3. Evaluate the square root:
[tex]\[
\sqrt{9} = 3
\][/tex]
4. Complete the equation with the substituted value:
[tex]\[
3 + 3 = 6
\][/tex]
5. Check if the simplified left side equals zero:
[tex]\[
6 \neq 0
\][/tex]
Since the left side of the equation simplifies to 6 and not 0, the equation [tex]\(\sqrt{2x + 1} + 3 = 0\)[/tex] does not hold true when [tex]\( x = 4 \)[/tex]. Hence, [tex]\( x = 4 \)[/tex] is not a solution to the equation.
By closely examining the algebraic steps, we can see that the student's conclusion was incorrect. The equation [tex]\(\sqrt{2x + 1} + 3 = 0\)[/tex] has no solution because the expression [tex]\(\sqrt{2x + 1} + 3\)[/tex] can never equal zero for any real value of [tex]\( x \)[/tex]. The square root function [tex]\(\sqrt{2x + 1}\)[/tex] always yields a non-negative result (i.e., [tex]\(\sqrt{2x + 1} \geq 0\)[/tex]), and adding 3 to a non-negative number will always give a result greater than or equal to 3, never zero. Therefore, there is no real value of [tex]\( x \)[/tex] that satisfies the initial equation.
