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

### 3 Answers | Add Yours

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.

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**

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}.**