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Please Help!!

Create two radical equations: one that has an extraneous solution, and one that does not have an extraneous solution. Use the equation below as a model:


a√(x+b) +c=d

User Edx
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1 Answer

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Given


a√(x+b)+c=d

we have


√(x+b)=(d-c)/(a)

Squaring both sides, we have


x+b=((d-c)^2)/(a^2)

And finally


x=((d-c)^2)/(a^2)-b

Note that, when we square both sides, we have to assume that


(d-c)/(a)>0

because we're assuming that this fraction equals a square root, which is positive.

So, if that fraction is positive you'll actually have roots: choose


a=1,\ b=0,\ c=2,\ d=6

and you'll have


√(x)+2=6 \iff √(x)=4 \iff x=16

Which is a valid solution. If, instead, the fraction is negative, you'll have extraneous roots: choose


a=1,\ b=0,\ c=10,\ d=4

and you'll have


√(x)+10=4 \iff √(x)=-6

Squaring both sides (and here's the mistake!!) you'd have


x=36

which is not a solution for the equation, if we plug it in we have


√(x)+10=4 \implies √(36)+10=4 \implies 6+10=4

Which is clearly false

User Lloyd Pique
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