Solve square root of (x+3)=x+1

√(x + 3)² = (x + 1)²..............square both sides to get rid of the radical, recall √N² = N
x + 3 = x² + 2x + 1..............(A + B)² = A² + 2AB + B²
x² + x - 2 = 0.......................get = 0
(x + 2)(x - 1) = 0
x + 2 = 0
x = -2
and
x- 1= 0
x = 1

square both sides

x + 3 = x^2 + 2x + 1
0 = x^2 + x - 2
0 = (x + 2) * (x - 1)
x = -2 , 1

Test

sqrt(-2 + 3) = -2 + 1
sqrt(1) = -2 + 1

If non-principal roots can be used, then x = -2 works, otherwise x = 1 is the only answer


x = 1
sqrt(1 + 3) = 1 + 1
sqrt(4) = 2
2 = 2

The sqrt(x+3) = x + 1 is fairly simple.

sqrt(x+3) = x + 1
sqrt(x+3) - 1 = x
x = 1

Check:
sqrt(x+3) = x + 1
sqrt(1+3) = 1 +1
sqrt(4) = 2

So it x = 1 checks.

Use the quadratic equation to solve this

OBVIOUSLY X IS 1

You might try x =1. Just a suggestion.