# What are the real solutions of equation square root (x+1) + square root( 2x+3) = 5 ?

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We have to solve sqrt (x+1) + sqrt ( 2x+3) = 5

sqrt (x+1) + sqrt ( 2x+3) = 5

square both the sides

x + 1 + 2x + 3 + 2*sqrt[(x + 1)(2x + 3)] = 25

=> 3x + 2*sqrt[(x + 1)(2x + 3)] = 21

=> 2*sqrt[(x + 1)(2x + 3)] = 21 - 3x

square both the sides

=> 4*(x + 1)(2x + 3) = (21 - 3x)^2

=> 4(2x^2 + 5x + 3) = 441 + 9x^2 - 126x

=> 8x^2 + 20x + 12 = 441 + 9x^2 - 126x

=> x^2 - 146x + 429 = 0

=> x^2 - 143x - 3x + 429 = 0

=> x(x - 143) - 3(x - 143) = 0

=> (x - 3)(x - 143) = 0

=> x = 3 and x = 143

**The solution of the equation is x = 3 and x = 143**

143 is not an answer only 3. You need to verify answers when using square roots, you cannot depend on intervals or just assuming something is an answer.

sqrt(3 + 1) + sqrt(2(3)+3) = 5

sqrt(4) + sqrt(9) = 5

2 + 3 = 5 checks....

sqrt(143 + 1) + sqrt(2(143)+3) = 5

sqrt(144) + sqrt(286+3) = 5

12 + sqrt(289) = 5

12 + 17 = 5 false so 143 is not a solution.

This is called an extraneous solution because sqrt(x) is the principal square root and is always positive. If we had +/- signs then 143 could be a solution.

before solving the equation, we'll impose the constraints of existence of square root:

x + 1 >= 0 => x >= -1

2x+ 3 >= 0

2x >= -3

x>= -3/2

The common interval of admissible values of x is [-1 , +infinite).

We'll solve the equation by raising to square both sides:

x+1+2x+3 + 2sqrt(x+1)(2x+3) = 25

3x + 4 + 2sqrt(x+1)(2x+3) = 25

2sqrt(x+1)(2x+3) = 21 - 3x:

2sqrt(x+1)(2x+3) = 3(7 - x)

We'll raise to square again:

4(x+1)(2x+3) = 9(7-x)^2

We'll expand the square and w'ell remove the brackets:

8x^2 + 20x + 12 = 441 - 126x + 9x^2

x^2 - 126x - 20x + 441 - 12 = 0

x^2 - 146x + 429 = 0

x1 = [146 + sqrt(21316 - 1716)]/2

x1 = (146 + 140)/2

x1 = 143

x2 = (146 - 140)/2

x2 = 3

**Since both values of x are in the common interval [-1 , +infinite), we'll validate them as solutions: {3 ; 143}.**