roots of equ

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rupsk: Master | Next Rank: 500 Posts; Posts: 123; Joined: Mon Feb 07, 2011 12:11 pmFollowed by: 1 member

by rupsk » Thu Aug 18, 2011 7:07 am

The condition that the roots of (x-a)/ (ax+1) = (x+b)/ (bx+1) are reciprocal to each other is

A. a - b = 0
B. b = 0
C. a = 0
D. a + b = 0
E. None of these

I picked B which is incorrect.

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Source: Beat The GMAT — Problem Solving | Thu Aug 18, 2011 7:07 am

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Anurag@Gurome: GMAT Instructor; Posts: 3835; Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members; Location: Milpitas, CA; Thanked: 1854 times; GMAT Score:770

by Anurag@Gurome » Thu Aug 18, 2011 7:14 am

rupsk wrote:The condition that the roots of (x-a)/ (ax+1) = (x+b)/ (bx+1) are reciprocal to each other is...

The equation is a quadratic equation and it will have two roots. If the roots are reciprocal to each other then the product of the roots will be 1.

--> (x - a)/(ax + 1) = (x + b)/(bx + 1)
--> (x - a)(bx + 1) = (x + b)(ax + 1)
--> (bxÂ² - abx + x - a) = (axÂ² + abx + x + b)
--> (a - b)xÂ² + 2abx + (a + b) = 0

Hence, product of the roots = (a + b)/(a - b)

If the roots are reciprocal to each other, (a + b)/(a - b) = 1
--> (a + b) = (a - b)
--> b = 0

The correct answer is B.

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rupsk: Master | Next Rank: 500 Posts; Posts: 123; Joined: Mon Feb 07, 2011 12:11 pmFollowed by: 1 member

by rupsk » Thu Aug 18, 2011 7:17 am

I did the same but i got wrong answer so thought of asking others if I had wrong approach or missed out in calculation. Once again thanks.

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