In the first problem, I’m going to assume that the square root is over the entire left side of the equation and not just the “2x”.

sqrt(2x+5) = x+1

2x+5 = (x+1)^2………………. (square both sides)

2x + 5 = x^2 + 2x + 1………. (multiply out the right side)

5 = x^2 + 1…………………….. (subtract 2x from both sides)

0 = x^2 – 4…………………….. (subtract 5 from each side)

x = 2 or -2…………………. (solve w/quadratic eq. or factors method)

To check the solution, plug 2 into the original equation and solve both sides (showing that you get 3 on both sides of the equation). Then plug -2 into the original equation and solve both sides (showing that you get different numbers on either side)… -2 is not a valid solution

For the second problem, I’m going to assume the first square root is over the entire left side, and the second is just over “x-2”

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

3x-2 = (sqrt(x-2) + 2)^2…………. (square both sides)

3x-2 = x + (4 * sqrt(x-2)) + 2 …. (multiply out the right side)

3x – 4 = x + (4 * sqrt(x-2))……… (subtract 2 from both sides)

2x – 4 = 4 * sqrt (x-2)……………. (subtract x from both sides)

x-2 = 2 * sqrt(x-2)…………………. (divide both sides by 2)

(x-2)^2 = 4(x-2)……………………. (square both sides)

x^2 – 4x +4 = 4x – 8………………. (multiply out both sides)

x^2 – 8x +4 = -8……………………. (subtract 4x from both sides)

x^2 – 8x + 12 = 0………………….. (add 8 to both sides)

x = 6 or 2…………… (solve using quadratic or factors method)

Again, to check, plug both answers into the original equation as you did above to make sure that you get a true equation. Disregard an answer if it doesn’t check out.

squaring both sides, we get

2x + 5 = (x+1)^2

2x + 5= x^2 + 2x + 1

x^2 – 4 = 0

(x-2) (x+2) = 0 [ using a^2 – a^2 = a-b x a+b]

x = 2, x = -2

Checking for x = 2

sqrt (2x+5) = x+1

substituting x = 2, we get

LHS: sqrt (2×2) +5 = sqrt (4 +5) = sqrt 9 = 3

RHS: 2 +1 = 3

Since LHS = RHS,

we can say that x = 2 is the solution to the equation sqrt (2x+5) = x+1

Checking for x = -2

sqrt (2x+5) = x+1

Substituting x = -2, we get

LHS: sqrt [(2x-2) + 5] = sqrt (-4 +5) = sqrt 1 = 1

RHS: -2 + 1 = -1

Since LHS is NOT equal to RHS,

therefore x = -2 is not the solution for the equation sqrt (2x+5) = x+1

FOr question 2… can u plz clearly mention what part is under square root.?

solution: x=2 or -2

The second one is equal to

x-2=2*(square root)x-2

or

3x¤2 -12x+12=0

with the only one solution x=2