Answer:
Given the equation: \(\sqrt{2x+1} =3\) ......[1]
Taking square both sides, we get;
\(2x+1 = 3^2\)
\(2x+1 = 9\)
Subtract 1 from both sides we get;
2x = 8
Divide both sides by 2 we get;
x = 4
Extraneous solution states that it is a root of a transformed equation that is not a root of the original equation because it was excluded from the domain of the original equation.
Substitute the value of x in [1] we get;
\(\sqrt{2(4)+1} =3\)
\(\sqrt{8+1}=3\)
\(\sqrt{9} =3\)
\(3 =3\) True.
Therefore, the value of x is 4 and it is not an extraneous solution.
