x2-5x-14An extraneous solution may occur when you're solving certain types of algebraic equation for the value of a variable. It's possible to arrive at more than one apparent solution (assuming you haven't made any mistakes in arriving at these solutions). However, when you carry out a check, using the original problem and substituting each of your solutions for the value of the variable, you discover that not all of the solutions satisfy the original equation. The solutions that don't satisfy the original equations are called extraneous solutions.
So what causes these extraneous solutions? A common cause is algebraic equations (or inequalities) which contain radicals, and the method for solving the equation involves squaring to get rid of the radicals. The process of squaring actually introduces an ambiguity and gives rise to an extraneous solution.
Another cause is an equation involving logarithms. You can only have logarithms of numbers exceeding zero, but, if one or more of the solutions is discovered to make the argument of a logarithm negative then these solutions would be invalid and therefore would be extraneous solution(s).
EXAMPLES
(1) √(x2+16x+16)=2x+1, squaring:
x2+16x+16=4x2+4x+1,
3x2-12x-15=0=3(x2-4x-5)=3(x-5)(x+1); x=5 or -1.
CHECK
x=5: √(25+80+16)=11✔️
x=-1: √(1-16+16)=-2+1=-1, but 1≠-1, so x=-1 is the extraneous solution, and x=5 is the only solution.
(2) log(x-7)+log(x+2)=log(x2-32)+log(2),
log[(x-7)(x+2)]=log[2(x2-32)],
log(x2-5x-14)=log(2x2-64), the log arguments must be equal, so
x2-5x-14=2x2-64,
x2+5x-50=0=(x+10)(x-5), so x=-10 or 5.
CHECK
x=-10: log(x-7)+log(x+2)=log(-17)+log(-8) which is meaningless because logs of negative numbers are not allowed.
x=5: log(x-7)+log(x+2)=log(-2)+log(7) which is also meaningless for the same reason. Also log(2x2-64)=log(-14).
Neither solution is valid, so both solutions are extraneous and there are no solutions.