Question

Simplify

sqrt(3 + 2 sqrt 2)

Answer 1

2.4141251

Answer 2

http://www.wolframalpha.com/input/?i=sqrt%283+%2B+2+sqrt+2%29
xDD

Answer 3

Null points...

Answer 4

simplify not calculate lol

Answer 5

Mind u he got a clue that way....

Answer 6

oh come on what's to simplyfy in this
a ninth std. student even knows how to calculate it

Answer 7

I'm the tutor and I want u to simplify it...:-)

Answer 8

2.4141251

Remind u of anything?

Answer 9

Calculate root 2

Answer 10

okay let me do it i didn't simplify it ah ...

Answer 11

Let's do algebra

Simplify sqrt(a+bsqrtc)

Answer 12

Stop alcatraz complaining....

Answer 13

ah i think I got it

Answer 14

sqrt( 1 + 2 +2sqrt2) = sqrt( (1+sqrt2)^2) = 1 + sqrt2

Answer 15

how's that

Answer 16

oh thanks : )

Answer 17

oh lols , ishaan pretending to be stupid xD

Answer 18

lol

Answer 19

Come on Ishaan, solve the general case....

Answer 20

ah okay

Answer 21

Ramanujan was good at these..

Answer 22

\[\sqrt{3+2*\sqrt{2}} = a + b*\sqrt{2}\]
Square both sides:
\[3+2*\sqrt{2} = (a+b*\sqrt{2})^2\]
\[3+2*\sqrt{2} = a^2 + 2b^2 + 2b*\sqrt{2}\]
Compare coefficients:
\[2 = 2b => b = 1\]
\[a^2 + 2b^2 = 3\]
\[a^2 + 2 = 3\]
\[a^2 = 1\]
\[a = \pm1\]
Therefore:
\[\sqrt{2} \pm 1\]

Should be correct :)

Answer 23

plus only, good work.

Answer 24

yep plus only, thanks :)

Answer 25

by the way there was a flaw in my working which made the previous answer coincidentally right :/
Here's the correct procedure:
\[\sqrt{3+2\sqrt{2}} = (a+b \sqrt{2})\]
Square both sides:
\[3+2 \sqrt{2} = (a+b \sqrt{2})^2\]\[3 + 2\sqrt{2} = a^2 + 2b^2 + 2ab \sqrt 2\]
Comparing Coefficients:
\[2ab = 2 => ab = 1\]\[a^2 + 2b^2 = 3\]
Solving simultanously:
\[b = {1 \over a}\]
therefore:
\[a^2 + 2({1 \over a^2}) - 3 = 0\]
multiply both sides by a^2:
\[a^4 - 3a^2 + 2 = 0\]
Let x = a^2, then:
\[x^2 - 3x + 2 = 0\]\[(x-1)(x-2) =0\]x = 1 or x = 2
a^2 = 1 or a^2 = 2
\[a = \pm 1 => b = \pm 1\]or\[a = \pm \sqrt{2} => b = \pm {\sqrt{2} \over 2}\]
Discarding negatives, one gets the pairs \[a = 1, b = 1\] or \[a = \sqrt{2}, b = {\sqrt{2} \over 2}\] both leading to the answer \[1 + \sqrt{2}\] when substituted into \[a + b \sqrt{2}\]

I apologize for the previous mistake, this is the correct working to the solution.