Simplifying An Interesting Radical

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[wagnerperlino]

The second method is much shorter. Congrats !

[dmitryli]

My solution was to inspect sqrt(x)+sqrt(x+2) in power of 2.... Once you open the parentheses, you get 2 of what you had in the original expression. That's about it

[josepherhardt]

Yeah, somebody pinch me. Actually worked this out ahead of viewing, in only about 5 lines of math.
And no, I did not think "Simplifying an Interesting Radical" was about analyzing Jerry Rubin or Timothy Leary. ;)

[bobbyheffley]

Negative x values yield imaginary numbers.

[aesthetics_ai]

i felt porud that i had the idea to make it in the beginning +1-1 to make it x^2 +2x +1 - 1 = (x+1)^2 - 1 then consider 1=1^2 thus making ( x+1)^2 - 1^2 = (x+1-1)(x+1+1) and it was all round to x^2 + 2x but yet good for a beginner i guess hehe

[Abdurhman_English]

The second method made my head spin 😅.

[Don-Ensley]

problem
Simplify
√[ x+1+√(x² +2x) ]

Let
√[ x+1+√(x² +2x) ] = √a + √b
Square.
x+1+√(x² +2x) = a+b+√(4ab)
Make a system of equations.
x + 1 = a + b
x² + 2 x = 4 a b

b = x + 1 - a
Substitute into second equation.
x² + 2 x = 4 a (x + 1 - a)

Solve for a.
x² + 2 x = 4 a x + 4 a - 4 a²
Rearrange into standard form as a quadratic in a.

4a²-4(x+1) a + x² + 2x = 0

Use the quadratic formula.

a = { 4(x+1) ± 4 } / 8
= { (x+1) ± 1 } / 2
= { (x+1) ± 1 } / 2

= (x + 2)/2, x / 2

b = x + 1 - a
= x + 1 -(x + 2)/2, x + 1 - x/2
= x / 2, ( x + 2 ) / 2

√[ x+1+√(x² +2x) ] = √a + √b

Since addition is symmetric, the radical simplifies to

√[ x+1+√(x² +2x) ] = √(x / 2) + √[(x + 2)/2]
= [√ x + √ (x+2)]/ √2

answer

√[ x+1+√(x² +2x) ] = √(x/2) + √(x/2+1)

[StaR-uwdc]

√(x+1+√(x²+2x)) = √(2x+2+2√(x(x+2)))/√2 = √(x+x+2+2√x√(x+2))/√2 = √(√x+√(x+2))²/√2 = (√x+√(x+2))/√2 = (√(2x)+√(2x+4))/2

[bayareapianist]

You wasted so much writing. All you had to do was to square both sides.