Let a and b be positive real numbers such that
1a−1b−1a+b=0
Find the value of (ba+ab)2.


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Answer:

5

Step by Step Explanation:
  1. We are given,
    1a−1b−1a+b=0
    Let x=ba
    ⟹b=ax
  2. Now,
    ⟹1a−1b−1a+b=0⟹1a−1ax−1a+ax=0⟹1a(1−1x−11+x)=0⟹1a(x(1+x)−(1+x)−xx(1+x))=0⟹1a(x+x2−1−x−xx(1+x))=0⟹1a(x2−x−1x(1+x))=0⟹x2−x−1x(1+x)=0(∵a>0)⟹x2−x−1=0⟹x=1±√1+42(1)⟹x=1±√52 Since a and b are positive, x > 0 ⟹x=1+√52
  3. Now,
    (ba+ab)=(x+1x)2=(1+√52+21+√5)2=(1+√52+21+√5×1−√51−√5)2=(1+√52+2−2√5−4)2=(1+√52−12+√52)2=(√5)2=5
  4. Hence, the value of (ba+ab)2 is 5.

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