7- sqrt. of x-6=3 (extraneous roots)?

Question

sqrt. of 5(x+3)=10

5sqrt of x-3=10

Answer 1

7 - √(x - 6) = 3 (extraneous roots)?

So √(x - 6) = 4
Square both sides
x - 6 = 16
x = 22

Check 7 - √(x - 6)
= 7 - √(22 - 6)
= 7 - 4
= 3 (No extraneous roots)

√5(x + 3) = 10

Square both sides
5(x + 3) = 100
x + 3 = 20
x = 17
Again on checking solution is fine

5√(x - 3) =10
So √(x - 3) = 2
square both sides
x - 3 = 7
x = 7

And again on checking solution is fine

Answer 2

7 - sqrt(x-6) = 3
- sqrt(x-6) = 3-7
- sqrt(x-6) = -4 (here you can either square this right away, or multiply all by -1, to turn them into positives, then square; we'll square it right away, because square of the number is the same regardless of its sign) /^2

x-6 = 4^2
x-6 = 16
x = 16+6
x = 22

check:
7 - sqrt(x-6) = 3
7 - sqrt(22-6) = 3
7 - sqrt(16) = 3
7 - sqrt(4^2) = 3
7 - 4 = 3
3 = 3
OK => x=22

sqrt. of 5(x+3)=10 /^2
(sqrt5(x+3)) = 10^2
5(x+3)=100 /:5
x+3 = 20
x = 20-3
x = 17

check:
sqrt. of 5(x+3)=10
sqrt. of 5(17+3)=10
sqrt(5*20)=10
sqrt(100) = 10
sqrt(10^2) = 10
10 = 10
OK => x = 17

5sqrt of x-3=10 /:5
sqrt(x-3)=2 /^2
x-3 = 4
x = 4+3
x = 7

check:
5sqrt of x-3=10
5sqrt(7-3)=10
5sqrt4=10
5sqrt(2^2)=10
5*2 = 10
10 = 10
OK => x = 7