A student concluded that the solution to the equation [tex]\sqrt{2x+1} + 3 = 0[/tex] is [tex]x = 4[/tex].

Do you agree? Explain why or why not.



Answer :

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.