how do you solve √(9+2√(20))?

9 + 2 rt( 20 )

= 5 + 4 + 2 rt( 5 x 4 )

= ( rt.5 )^2 + 2^2 + 2.( rt.5).2

= ( rt.5 + 2 )^2

Hence, the req'd sqrt = +/- ( rt.5 + 2 )............Ans.

Simplify √[9 + 2√(20)].

Square the following:
(√a + √b)² = a + 2√(ab) + b = (a + b) + 2√(ab)

So we need:
a + b = 9 and ab = 20

The solution is:
a = 4 and b = 5
_________

√[9 + 2√(20)] = √(√4 + √5)² = √4 + √5 = 2 + √5

well √(20) = √(4*5) = √(4)√(5)=2√(5)
so √(9+2√(20)) = √(9+4√(5))

Let (x+y√(5)) = √(9+4√(5)) where x and y are rational (no square root)

therefore (x+y√(5))^2 = 9+4√(5)

x^2 + 2xy√5 + 5y^2 = 9 + 4√(5)

Therefore x^2 + 5y^2 = 9 (no square root in the term)
2xy = 4 (terms with square root in the term)
xy = 2

Solving simultaneously:
Substitute y = 2/x into x^2 + 5y^2 = 9
x^2 + 5(2/x)^2 = 9
x^2 + 20/x^2 = 9
x^4 + 20 = 9x^2
x^4 - 9x^2 + 20 = 0
x^4 - 4x^2 - 5x^2 + 20 = 0
x^2(x^2 - 4) - 5(x^2 - 4) = 0
(x^2 - 4)(x^2 - 5) = 0
so x^2 - 4 = 0 or x^2 - 5 = 0

x^2 = 4 or x^2 = 5
x = +/- 2 or x = +/- √(5)
but we only want rational x
so x = +/- 2
As:
xy = 2
y = +/- 1
But x+y√(5) = √(9+4√(5))
so √(9+4√(5)) = -2 + (-1)√(5) = -2 -√(5) bu this isn't equal to √(9+4√(5))
so √(9+4√(5)) = 2 + 1√(5) = 2 + √(5).

so √(9+2√(20)) = 2 + 1√(5) = 2 + √(5).

Let x=√(9+2√(20))

= sqrt(9 +2 * 4.4721)
= sqrt(9+ 8.9443)
= sqrt ( 17.9443)
=4.236
***
√(9+2√(20))
=√(9+2√(4x5))
=√(9+2x2√(5))
=√(9+4√(5))

You're not solving it, you're evaluating it (to solve something you need a left hand side and a right hand side and an equals sign in the middle).

It's √(9 + 4√5) which is roughly 4.23607

solve wht is under the first radicand then go from there... :D

√ [ 9 + 2 √20 ]

√ [ 9 + 4√5 ]

9+2√(20) = 5 + 2√20 + 4 = (√5)^2 +2√(5)√(4) +(√4)^2
= (√5 + √4)^2 => √(9+2√(20)) = √5 + √4

√(9 + 2√20)
= √[9 + 2√(2^2 * 5)]
= √[9 + 2(2)√5]
= √[9 + 4√5]
= 4.23606798

hmmmm