Question

(a+b)^2=400 and ab=64 then the value of a and b

Answer 1

( a + b )^2 = 400
=> a^2 + b^2 + 2ab = 400
=> a^2 + b^2 + 2ab + 2ab - 2ab = 400
=> (a^2 + b^2 - 2ab) + (2ab + 2ab) = 400
=> (a - b)^2 + 4ab = 400
=> (a - b)^2 + 4 x 64 = 400
=> (a - b)^2 + 256 = 400
=> (a - b)^2 = 400 - 256
=> (a - b)^2 = 144
=> Root { (a - b)^2 } = Root ( 144 )
=> a - b = 12 ---- ( i )

Also we have
( a + b )^2 = 400
=> Root ( ( a + b )^2) = Root ( 400 )
=> a + b = 20 ----( ii )

from equ. ( i ) , a = 12 + b ----( iii ) , putting the value of 'a' in equ.(ii) gives ,
12 + b + b = 20
=> 12 + 2b = 20
=> 2b = 20 -12
=> 2b = 8
=> b = 4 , putting in equ.(iii) gives ,
a = 12 + 4 = 6
Thus a = 6 and b = 4