Lets solve our radical equation [tex] \sqrt{x+6}-4=x [/tex] step by step.
Step 1 add 4 to both sides of the equation:
[tex] \sqrt{x+6} -4+4=x +4[/tex]
[tex] \sqrt{x} +6=x+4 [/tex]
Step 2 square both sides of the equation:
[tex] \sqrt{x+6} =x+4 [/tex]
[tex] \sqrt{x+6}^2 =(x+4)^2 [/tex]
[tex] x+6=(x+4)^2 [/tex]
Step 3 expand the binomial in the right hand side:
[tex] x+6=x^2+8x+16 [/tex]
Step 4 simplify the expression:
[tex] 0=x^2+8x-x+16-6 [/tex]
[tex] x^2+7x+10=0 [/tex]
Step 5 factor the expression:
[tex] (x+2)(x+5)=0 [/tex]
Step 6 solve for each factor:
[tex] x+2=0 [/tex] or [tex] x+5=0 [/tex]
[tex] x=-2 [/tex] or [tex] x=-5 [/tex]
Now we are going to check both solutions in the original equation to prove if they are valid:
For [tex] x=-2 [/tex]
[tex] \sqrt{x+6}-4=x [/tex]
[tex] \sqrt{-2+6}-4=-2 [/tex]
[tex] \sqrt{4}-4=-2 [/tex]
[tex] 2-4=-2 [/tex]
[tex] -2=-2 [/tex]
The solution [tex] x=2 [/tex] is a valid solution of the rational equation [tex] \sqrt{x+6}-4=x [/tex].
For [tex] x=-5 [/tex]
[tex] \sqrt{x+6}-4=x [/tex]
[tex] \sqrt{-5+6}-4=-5 [/tex]
[tex] \sqrt{1}-4=-5 [/tex]
[tex] 1-4=-5 [/tex]
[tex] -3\neq -5 [/tex]
Since -3 is not equal to -5, the solution [tex] x=-5 [/tex] is not a valid solution of the rational equation [tex] \sqrt{x+6}-4=x [/tex]; therefore, [tex] x=-5 [/tex] is an extraneous solution of the equation.
We can conclude that even all the algebraic procedures of Israel are correct, he did not check for extraneous solutions.
An extraneous solution of an equation is the solution that emerges from the algebraic process of solving the equation but is not a valid solution of the equation. Is worth pointing out that extraneous solutions are particularly frequent in rational equation.
