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. [tex] \sqrt{12-(-4)} = \sqrt{16} = \frac{+}{}4[/tex] Extraneous root in this case is positive four since +4≠-4 [tex] \sqrt{12-3} = \sqrt{9} = \frac{+}{}3[/tex] In this case it's negative 3, since -3≠3
#29 can be turned into a quadratic equation. [tex]x= \sqrt{2x+3}[/tex] Square both sides to get [tex]x^{2}=2x+3[/tex] Then bring the 2x+3 to the other side, setting the quadratic equal to zero. [tex] x^{2}-2x-3=0[/tex] 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. [tex]x= \sqrt{2x+3}[/tex] [tex]3= \sqrt{2(3)+3}[/tex] [tex]3= \sqrt{9}[/tex] Extraneous root would be negative 3.
[tex]-1= \sqrt{2(-1)+3}[/tex] [tex]-1= \sqrt{1}[/tex] Extraneous root would be positive 1.
Your answers are positive 3 and negative 1. Extraneous roots are negative 3 and positive 1.
