Question

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

Answer 1

#25 is fairly simple. Plug in -4 and 3 into the equation, and the extraneous root will be the one that does not work.

`√(12-(-4)) = √(16) = \frac{+}{}4`
Extraneous root in this case is positive four since +4≠-4

`√(12-3) = √(9) = \frac{+}{}3`
In this case it's negative 3, since -3≠3

#29 can be turned into a quadratic equation.

`x= √(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= √(2x+3)`

`3= √(2(3)+3)`

`3= √(9)`
Extraneous root would be negative 3.


`-1= √(2(-1)+3)`

`-1= √(1)`
Extraneous root would be positive 1.

Your answers are positive 3 and negative 1.
Extraneous roots are negative 3 and positive 1.