x = -3
sqrt(3 - 3) - 1 =>
0 - 1 =>
-1
sqrt(3 * -3 + 10) =>
sqrt(-9 + 10) =>
sqrt(1) =>
-1 or 1.
x = -3 CAN work, but ONLY if we're allowed to use negative roots. If not, then x = -3 is an extraneous answer
x = -2
sqrt(3 - 2) - 1 =>
+/- 1 - 1 =>
-2 , 0
sqrt(3 * (-2) + 10) =>
sqrt(-6 + 10) =>
sqrt(4) =>
-2 , 2
Once again, if negative roots are permitted, then x = -2 is an answer. If not, then it is extraneous.
But if both answers are extraneous, then what does that mean? That there isn't an answer.
Now, do you remember that I said I could use substitution to solve? Would you like to see how?
-sqrt(3 + x) = x + 3
Let x + 3 = t
-sqrt(t) = t
Let t^(1/2) = u
-u = u^2
0 = u^2 + u
0 = u * (u + 1)
Using the Zero-Product property, we can split this apart (if a * b = 0, then a = 0 or b = 0 , or both a and b are equal to 0)
u = 0
u + 1 = 0
u = -1
u = 0 , -1
u = t^(1/2)
Let's back-substitute
t^(1/2) = 0 , -1
t = 0^2 , (-1)^2
t = 0 , 1
x + 3 = 0 , 1
x = 0 - 3 , 1 - 3
x = -3 , -2
Same possible answers, a slightly different method.