Pretty sure you already had something posted about this... in case you've lost it, or have questions:
#25 is fairly simple. Plug in -4 and 3 into the equation, and the extraneous root will be the one that does not work.
\(\sqrt{12-(-4)} = \sqrt{16} = \frac{+}{}4\)
Extraneous root in this case is positive four since +4≠-4
\(\sqrt{12-3} = \sqrt{9} = \frac{+}{}3\)
In this case it's negative 3, since -3≠3
#29 can be turned into a quadratic equation.
\(x= \sqrt{2x+3}\)
Square both sides to get
\(x^{2}=2x+3\)
Then bring the 2x+3 to the other side, setting the quadratic equal to zero.
\(x^{2}-2x-3=0\)
Factor to find that it's equivalent to
(x-3)(x+1)=0
Therefore x is equal to positive 3 and negative 1. Plug both back into the original equation. Whichever does not work is the extraneous root, and the answer is the one that does.
\(x= \sqrt{2x+3}\)
\(3= \sqrt{2(3)+3}\)
\(3= \sqrt{9}\)
Extraneous root would be negative 3.
\(-1= \sqrt{2(-1)+3}\)
\(-1= \sqrt{1}\)
Extraneous root would be positive 1.
Your answers are positive 3 and negative 1.
Extraneous roots are negative 3 and positive 1.
