Sally found that a solution of x = 2 was extraneous for her function. What would be a possible denominator in Sally's function? 4x + 8 2x + 12 x2 - 4 x2 - 16
@kritikachandak
extraneous is bad. example: if x=3 is an extraneous solution, we might have a denominator of:\(\Large\rm x-3\)
extraneous means its not relevant to the equation right
This denominator gives us a zero in the denominator when x=3. So it's bad!:O yah, it's not a workable solution. we can ignore it as a solution.
Since we have x=2 as extraneous, (x-2) is the form we're looking for in the denominator. Do any of our options give us that? We'll have to factor to see correctly.
ok so how do we find the denominator?
x=2 is extraneous, \(\Large\rm \color{red}{(x-2)}\) is the factor we'e looking for. \(\Large\rm 4x+8=4(x+2)\) Hmm that one didn't give us a factor of x-2. So it's not option A.
Understand how that works? :o
yessir
ahh sorry class is starting :c i gotta go
:( darnnit
\(\Large\rm x^2-4=(x+2)\color{red}{(x-2)}\)
is it x^2-4 real quick?
Remember how to factor the difference of squares?
oh boy the website is tweaking out on me :c
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