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sqrt(x + 1) = 5 - sqrt(x - 4) square both sides. YOu could do it your way. It's going to come to the same thing. This is the way I do it. 6 to one 1/2 dozen to the other.

x + 1 = 25 - 10*sqrt(x - 4) + x - 4 Now transfer the some of right side to the left side.

x - x + 1 - 25 +4 = - 10*sqrt(x - 4)

- 20 = - 10* sqrt(x - 4) multiply by - 1

20 = 10 sqrt(x - 4)

2 = sqrt(x - 4) square both sides

4 = x - 4

x = 8

Converting to symmetrical forms can get rid of radicals quite elegantly.

Let p = x - 1.5 (half-way between x+1 and x-4)

Then you have:

√(p + 2.5) + √(p - 2.5) = 5

√(2p + 5) √(2p - 5) + 2p = 25 ... square both sides

√((4p^2 - 25) = 25 - 2p ... multiply and rearrrange

4p^2 - 25 = 4p^2 - 100p + 625 ... square both sides again

- 25 = - 100p + 625 ... simplify

100p = 650

p = 6.5

x = p + 1.5 = 8 ... reverse substitution

√(8+1) + √(8-4) = √9 + √4 = 3 + 2 = 5 QED

I feel this is not terribly rigorous, but the equation is undefined in reals for x<4, and a graph confirms that this is the only real solution.

That's how I would do it.

√(x+1) + √(x-4) = 5

(x+1)+(x-4)=5^2

(x+1)+(x-4)=25

x+1+x-4=25

2x-3=25

2x=28

x=14

Obviously the answer is 8 and not 14 if you check it......I'm just not sure how to solve and get 8. I will look around.

√(x + 1) + √(x - 4) = 5

√(x + 1) = 5 - √(x - 4)

(x + 1) = 25 - 10√(x - 4) + (x - 4)

5 = 25 - 10√(x - 4)

2 = √(x - 4) => x = 8

Check:

√(x + 1) + √(x - 4) = √9 + √4 = 3 + 2 = 5

√(x + 1) + √(x - 4) = 5

[√(x + 1) + √(x - 4)]^2 = 5^2

x + 1 + 2√(x + 1)√(x - 4) + x - 4 = 25

2x + 2√(x^2 - 3x - 4) = 28

2√(x^2 - 3x - 4) = 28 - 2x

√(x^2 - 3x - 4) = 14 - x

x^2 - 3x - 4 = (14 - x)^2

x^2 - 3x - 4 = 196 - 28x + x^2

25x = 200

x = 8