Question

Please Help!!

Answer 1

Given

$a\sqrt{x+b}+c=d$

we have

$\sqrt{x+b}=\dfrac{d-c}{a}$

Squaring both sides, we have

$x+b=\dfrac{(d-c)^2}{a^2}$

And finally

$x=\dfrac{(d-c)^2}{a^2}-b$

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

$\dfrac{d-c}{a}>0$

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

$\sqrt{x}+2=6 \iff \sqrt{x}=4 \iff x=16$

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

$\sqrt{x}+10=4 \iff \sqrt{x}=-6$

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

$\sqrt{x}+10=4 \implies \sqrt{36}+10=4 \implies 6+10=4$

Which is clearly false