Cracking The SAT Premium

Once simplified, two of the three denominators become zero. That is not allowed, so the solution you
found isn’t really a solution at all. It is referred to as an “extraneous solution.” That term refers to any
answer you get to an algebraic equation that results in a false statement when plugged back in to the
original equation.

Here’s another example.

29.

    =   t   –   2

In  the equation    above,  what    is  the value   of  the extraneous  solution,   if  one exists?

A) 0

B) 4

C) 5

D) There    are no  extraneous  solutions.

Extra   Answers

Any time    you are solving

for a   variable,   make    sure

your    solutions   actually

work.   If  they    do  not,    they

are “extraneous,”   or  extra.

Here’s How to Crack It

Start by squaring both sides of the equation to get rid of the radical. The equation becomes

t   +   4   =   (t  –   2)^2

Use FOIL to multiply the right side of the equation to get t^2 – 2t – 2t + 4 or t^2 – 4t + 4. Now the equation
is

t   +   4   =   t^2     –   4t  +   4

Subtract t and 4 from both sides to get

0   =   t^2     –   5t

The right side factors to t(t – 5), so t = 0 or 5. Eliminate (B), since 4 is not a solution at all, extraneous or
otherwise. Now plug 0 and 5 back into the original equation to see if they work. If both do, the answer is