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:[tex]a\sqrt{x+b} +c=d[/tex]

Given[tex]a\sqrt{x+b}+c=d[/tex]we have[tex]\sqrt{x+b}=\dfrac{d-c}{a}[/tex]Squaring both sides, we have[tex]x+b=\dfrac{(d-c)^2}{a^2}[/tex]And finally[tex]x=\dfrac{(d-c)^2}{a^2}-b[/tex]Note that, when we square both sides, we have to assume that[tex]\dfrac{d-c}{a}>0[/tex]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[tex]a=1,\ b=0,\ c=2,\ d=6[/tex]and you'll have[tex]\sqrt{x}+2=6 \iff \sqrt{x}=4 \iff x=16[/tex]Which is a valid solution. If, instead, the fraction is negative, you'll have extraneous roots: choose[tex]a=1,\ b=0,\ c=10,\ d=4[/tex]and you'll have[tex]\sqrt{x}+10=4 \iff \sqrt{x}=-6[/tex]Squaring both sides (and here's the mistake!!) you'd have[tex]x=36[/tex]which is not a solution for the equation, if we plug it in we have[tex]\sqrt{x}+10=4 \implies \sqrt{36}+10=4 \implies 6+10=4[/tex]Which is clearly false