Remember that an extraneous solution of an equation is a solution that emerges from the algebraic process of solving the equation but is not a valid solution of the equation.
First, we are going to solve our equation algebraically:
Step 1 simplify the equation:
\(\sqrt{x} +9-4=1\)
\(\sqrt{x}+ 5=1\)
Step 2 subtract 5 from both sides of the equation:
\(\sqrt{x} +5-5=1-5\)
\(\sqrt{x} =-4\)
Step 3 square both sides of the equation:
\(\sqrt{x} ^2=(-4)^2\)
\(x=16\)
Next, we are going to replace our solution in our original equation and check if it is a valid solution:
\(\sqrt{x} +9-4=1\)
\(\sqrt{16} +5=1\)
\(4+5=1\)
\(9
eq 1\)
Since 9 is not equal to 1, \(x=16\) is not valid solution of the equation; therefor it is an extraneous solution.
We can conclude that the correct answer is: x = 16, solution is extraneous
