Question

\[\frac { \sqrt { 2 } -1 }{ \sqrt { 2 } +1 } =\quad a+b\sqrt { 2 } \quad then\quad find\quad the\quad value\quad of\quad a\quad and\quad b. \]

Answer 1

Can ANyONe SOlVe It??

Answer 2

@UnkleRhaukus

Answer 3

ok lets look at the left hand side first

Answer 4

okay! carry on!!

Answer 5

\[LHS=\frac { \sqrt { 2 } -1 }{ \sqrt { 2 } +1 } \]
\[=\frac { \sqrt { 2 } -1 }{ \sqrt { 2 } +1 } \times1\]
\[=\frac { \sqrt { 2 } -1 }{ \sqrt { 2 } +1 } \times\frac{\sqrt 2-1}{\sqrt 2-1}\]

Answer 6

simplify

Answer 7

hmm....

Answer 8

Hey always if the denominator is a number, then it is for us to solve it. So first try to do that. If a square root exists in the denominator try to make it a number. Now think we've a sqrt her...so shd square it without changing the fraction......option is that use (a+b)(a-b)=aa-b2

Answer 9

\[=\frac {( \sqrt { 2 } -1)(\sqrt 2-1) }{ (\sqrt { 2 } +1 )(\sqrt 2-1)} \]

Answer 10

say a+b is there in the denominator, multiply the fraction i.e both numerator and denom woth a-b. So in denom it'll be a2-b2......Solved

Answer 11

ok

Answer 12

hmmmmm...

Answer 13

what are you getting for the numerator and denominator ?

Answer 14

m confused

Answer 15

\[( \sqrt { 2 } -1)(\sqrt 2-1) =\]

\[{ (\sqrt { 2 } +1 )(\sqrt 2-1)}=\]

Answer 16

A^2- b^2

Answer 17

UNKLE GIVE the solution

Answer 18

Thanks mathy

Answer 19

got it ? ?

Answer 20

thanks

Answer 21

given is in the form (a-b)/(a+b).....
multiply with a-b to numerator & denom
it becomes (a-b)2/(a2-b2)........read a2=Square of a
substitute the values of a and b in this problem to get the answer

Answer 22

welcome

Answer 23

Indian gods

Answer 24

Correct.... not Indian Gods but only Gods. These gods an worshipped all over world

Answer 25

cool!!

Answer 26

@mathslover how a= 3 &b = -2

Answer 27

compare both eqn.