Arun
Last Activity: 7 Years ago
Dear Sunny
Given a + √b = c + √d
Case (i): Let a=c
⇒ a + √b = c + √d becomes
a + √b = a + √d
⇒ √b = √d
∴ b = d
Case (ii): Let a ≠ c
Let us take a = c + k where k is a rational number not equal to zero.
⇒ a + √b = c + √d becomes
(c + k) + √b = c + √d
⇒ k + √b = √d
Let us now square on both the sides,
⇒ (k + √b)2 = (√d)2
⇒ k2 + b + 2k√b = d
⇒ 2k√b = d – k2 – b
sqrt(b) = [d - k² - b]/2k
Notice that RHS is a rational number.
Hence sqrt(b) is a rational number.
This is possible when b is a square of rational number.
Thus d is also a square of rational number as
k + sqrt(b) = sqrt (d)
Regards
Arun (askIITians forum expert)
