Question

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`

Answer 1

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