Given the equation Square root of 2x plus 1 = 3, solve for x and identify if it is an extraneous solution. x = 4, solution is extraneous x = 4, solution is not extraneous x = 5, solution is extraneous x = 5, solution is not extraneous

[tex]\text{The domain:}\\2x+1\geq0\to 2x\geq-1\to x\geq-0.5\\\\D:x\geq0.5\to x\in\left\ \textless \ 0.5;\ \infty\right)\\\\\sqrt{2x+1}=3\ \ \ \ |\text{square both sides} \\2x+1=3^2\\2x+1=9\ \ \ |-1\\2x=8\ \ \ \ |:2\\x=4\in D\\\\\text{Answer: x =0, solution is not extraneous}.[/tex]

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ColinJacobus ColinJacobus · Answer 2 · 2019-06-22T10:29:00+02:00

Answer: The correct option is

(B) solution is x = 4, the solution is not extraneous.

Step-by-step explanation: We are given to solve the following equation :

[tex]\sqrt{2x+1}=3~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]

And, we are to select the correct option about the solution of the above equation.

From equation (i), we have

[tex]\sqrt{2x+1}=3\\\\\Rightarrow 2x+1=3^2~~~~~~~~~~~~~~[\textup{Squaring both sides}]\\\\\Rightarrow 2x+1=9\\\\\Rightarrow 2x=9-1\\\\\Rightarrow 2x=8\\\\\Rightarrow x=\dfrac{8}{2}\\\\\Rightarrow x=4.[/tex]

So, x = 4 is a solution of the given equation.

Now, substituting x = 4 in the left hand side of equation (i), we get

[tex]L.H.S.=\sqrt{2\times4+1}=\sqrt{8+1}=\sqrt{9}=3=R.H.S.[/tex]

Since x = 4 satisfy the given equation, so it is not an extraneous solution.

Thus, the required solution is x = 4, the solution is not extraneous.

Option (B) is CORRECT.