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Kel Sir Solution 1

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Sushant
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23 views2 pages

Kel Sir Solution 1

Uploaded by

Sushant
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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Solution

Given the equations:


a √
a+ (Eqn 1)
a=
b
√ b
b+ b= (Eqn 2)
a
√ √
Since a ̸= 0, b ̸= 0, and a > 0, b > 0, we proceed as follows.
Multiply both sides of Eqn (1) by b:
√
ab + b a = a (Eqn 3)

Multiply both sides of Eqn (2) by a:


√
ab + a b = b (Eqn 4)

Subtract Eqn (4) from Eqn (3):


√ √
b a−a b=a−b

Factorize: √ √ √ √ √ 
( a − b) a + b + ab = 0
√ √ √
Since a+ b+ ab > 0, it follows that:
√ √
a= b

a = b (Eqn 5)
Substitute a = b into Eqn (1):
√
a+ a=1
√
Let a = k, so:
k2 + k = 1
Solve the quadratic equation:

k2 + k − 1 = 0
√
−1 ± 5
k=
2
Since k > 0, we take: √
−1 + 5
k=
2
Thus: √
√ −1 + 5
a=
2

1
√ !2 √
−1 + 5 3− 5
a= =
2 2
Since a = b, we have:
√ √ !
3− 5 3− 5
(a, b) = ,
2 2

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