Radical equation?

Question

Square root (2x+3)- square root(x+1) =1


I don't know how to enter square root symbol on the computer, so equation in ( ) is supposed to be under sq rt symbol.

why do you square both sides twice?

Answer 1

√(2x+3) - √(x+1) = 1
Square both sides
2x+3 - 2√(2x+3)√(x+1) + x+1 = 1
Regroup:
2√(2x+3)√(x+1) = 3x + 3
Square both sides:
4(2x+3)(x+1) = 9x^2 + 18x + 9
Expand:
8x^2 + 20x + 12 = 9x^2 + 18x + 9
Regroup:
x^2 - 2x - 3 = 0
(x-3)(x+1) = 0

x = 3 or x = -1

Its always a good idea with radical equations to check that every solution satisfies the original equation, as there is often the possibility of extraneous roots. Both of these solutions do happen to satisfy the original equation.

*************
Edit after Additional Details:
Notice that squaring once will eliminate one radical, whether you square the original equation as I did above, or you isolate one radical on one side of the equation and then square as others have done (either way works). One radical term will then remain. {Note above that the term 2√(2x+3)√(x+1) remains after one squaring. This term is really just one radical since:
2√(2x+3)√(x+1) = 2√[(2x+3)(x+1)] }. After the first squaring you isolate the remaining radical on one side of the equation and then square again, thus eliminating that remaining radical.

Answer 2

1. Square both sides of the equation. That will leave you with a new equation that contains a product of two radicals rather than a difference of two radicals.

2. Combine the product of the radicals, so that you have just a single radical.

3. Rearrange the equation so that the (single) radical is by itself on one side of the equation.

4. Square both sides again. This will get rid of the last remaining radical.

5. You should now have an equation in terms of x² and x. This is a quadratic equation. Combine all the like terms, then use the quadratic formula to solve.

Answer 3

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

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

squaring on both sides

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

x + 1= 2sqrt(x+1)

again squaring

x^2 + 1 + 2x = 4(x+1) = 4x + 4

x^2 -2x - 3 = 0

(x - 3)(x+1) = 0

x = 3 or - 1

Answer 4

sqrt ( 2x+3) - sqrt (x+1) = 1
sqrt ( 2x+3) = 1 + sqrt (x+1)
squaring both sides,
2x+3 = 1 + 2*sqrt(x+1) + (x+1)
x+1 = 2*sqrt(x+1)
square both sides again,
x^2 + 2x + 1 = 4(x+1)
x^2 + 2x + 1 = 4x + 4
x^2 -2x -3 = 0
factoring,
(x-3)*(x+1) = 0
x = 3 or x = -1