Answer:
The correct option is 1.
Step-by-step explanation:
The given equation is
\(\sqrt{4x+41}=x+5\)
Taking square on both the sides.
\(4x+41=(x+5)^2\)
\(4x+41=x^2+10x+25\) \([\because (a+b)^2=a^2+2ab+b^2]\)
\(0=x^2+10x+25-4x-41\)
\(0=x^2+6x-16\)
\(0=x^2+8x-2x-16\)
\(0=x(x+8)-2(x+8)\)
\(0=(x+8)(x-2)\)
\(x=2,-8\)
Check the equation by x=2.
\(\sqrt{4(2)+41}=2+5\)
\(\sqrt{49}=7\)
\(7=7\)
LHS=RHS, therefore x=2 is not an extraneous solution.
Check the equation by x=-8.
\(\sqrt{4(-8)+41}=-8+5\)
\(\sqrt{9}=-3\)
\(3=-3\)
LHS≠RHS, therefore x=-8 is an extraneous solution.
Therefore option 1 is correct.
