Question

SEND THE PHOTO OF YOUR SOLUTION.

Answer 1

HELLO DEAR,

if you want to prove this

now,

condition --(1)

let, a = c

we get,

a+√b = a+ √d

=> √b = √d
=> b = d
---------------or ---------------

c+ √b = c +√d

=> √b = √d

=> b = d


condition---(2)

let a ≠ c

TAKE, a = c + X

where ,

X is a rational number not equal to zero.

=> a + √b = c + √d

=> (c + X) + √b = c + √d

=> X + √b = √d
[ NOW SQUARING ON BOTH THE SIDES]

=> X² + b +2X√b = d

=>
$2x \sqrt{b} = d - {x}^{2} - b \\ = > \sqrt{b} = \frac{(d - {x}^{2} - b)}{2x}$
we know that:-

d , b , X are rational number

therefore b is a square of rational no.

X+√b =√d

=> √d is a rational no.

d is the square of rational number.



$I HOPE ITS HELP YOU DEAR, \\ THANKS$

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