Question

I am confused on numbers 25 and 29, the instructions are at the top.

Answer 1

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.