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…

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asked Jan 8, 2020 232k views

1 Answer

Lloyd Pique · Answer 1 · 2020-01-12T00:30:23+0000

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