Question

If x1=√a+3b+√a−3b√a+3b−√a−3b then, which one of the following statements is true?

A

3bx2+9b2ax=0

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B

3bx2+3b2ax=0

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C

3bx2+3b6ax=0

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D

5bx2+3b2ax=0

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Solution

The correct option is B 3bx2+3b−2ax=0 x1=√a+3b+√a−3b√a+3b−√a−3b Applying componendo and dividendo x+1x−1=√a+3b+√a−3b+√a+3b−√a−3b√a+3b+√a−3b−√a+3b+√a−3b = 2√a+3b2√a−3b = √a+3b√a−3b Squaring on both sides, we get: (x+1)2(x−1)2=(√a+3b)2(√a−3b)2 = (x+1)2(x−1)2=a+3ba−3b Again, applying componendo and dividendo (x+1)2+(x−1)2(x+1)2−(x−1)2=a+3b+a−3ba+3b−a+3b 2(x2+1)4x=2a6b =(x2+1)2x=a3b =3bx2+3b=2ax = 3bx2+3b−2ax=0

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