Question

Solve it

`√(2x + 1) - √(x + 1) = 2`
check for extraneous solutions​

Answer 1

Take note of the domain of possible solutions:

• √(2x + 1) is defined for 2x + 1 ≥ 0, or x ≥ -1/2

• √(x + 1) is defined for x + 1 ≥ 0, or x ≥ -1

So any solutions to this equation must fall in the interval x ≥ -1/2, or [-1/2, ∞).

To solve, move one of the root expressions to the right side, then square both sides:

√(2x + 1) = 2 + √(x + 1))

[√(2x + 1)]² = [2 + √(x + 1))]²

2x + 1 = 4 + 4 √(x + 1) + (x + 1)

Simplify this to

x - 4 = 4 √(x + 1)

and now square both sides again and solve for x :

[x - 4]² = [4 √(x + 1)]²

x ² - 8x + 16 = 16 (x + 1)

x ² - 8x + 16 = 16x + 16

x ² - 24x = 0

x (x - 24) = 0

x = 0 or x - 24 = 0

x = 0 or x = 24

Both of these solutions are greater than -1/2; however, if x = 0, we get

√(2×0 + 1) - √(0 + 1) = √1 - √1 = 1 - 1 = 0 ≠ 2

so the only solution is x = 24.

Answer 2

• √(2x + 1) is defined for 2x + 1 ≥ 0, or x ≥ -1/2

• √(x + 1) is defined for x + 1 ≥ 0, or x ≥ -1

So any solutions to this equation must fall in the interval x ≥ -1/2, or [-1/2, ∞).

To solve, move one of the root expressions to the right side, then square both sides:

√(2x + 1) = 2 + √(x + 1))

[√(2x + 1)]² = [2 + √(x + 1))]²

2x + 1 = 4 + 4 √(x + 1) + (x + 1)

Simplify this to

x - 4 = 4 √(x + 1)

and now square both sides again and solve for x :

[x - 4]² = [4 √(x + 1)]²

x ² - 8x + 16 = 16 (x + 1)

x ² - 8x + 16 = 16x + 16

x ² - 24x = 0

x (x - 24) = 0

x = 0 or x - 24 = 0

x = 0 or x = 24

Both of these solutions are greater than -1/2; however, if x = 0, we get

√(2×0 + 1) - √(0 + 1) = √1 - √1 = 1 - 1 = 0 ≠ 2

so the only solution is x = 24.