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kindly explain me how to solve it............

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by sheelanadh » Mon Aug 09, 2010 8:57 pm
The equation formed by decreasing each root of ax^2+bx+c=0 by 1 is 2x^2+bx+2=0, then:

(a) a=-b
(b) b=-c
(c) c=-a
(d) b=a+c
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by 4GMAT_Mumbai » Mon Aug 09, 2010 9:16 pm
Hi, Interesting q ... Thanks.

I thought the best ways was to assume values for b in the 2nd equation

Let b = -5; The equation becomes 2x^2-5x+2=0

Roots are 1/2 or 2.

Roots of 1st equation are 3/2 or 3

try to form the 1st equation

(x-3/2) (x-3) = 0

Make sure that the coefficients of x match in both the equation. (as they are equal to b).

Only one of the answer choices meet the criteria ... May I reserve the answer at this stage ...

Look forward to know better methods
Naveenan Ramachandran
4GMAT, Dadar(W) & Ghatkopar(W), Mumbai
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by selango » Mon Aug 09, 2010 11:25 pm
sheelanadh,

Is this a GMAT question?It's seems to be CAT question..
--Anand--
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by likithae » Mon Aug 09, 2010 11:35 pm
4GMAT_Mumbai wrote:Hi, Interesting q ... Thanks.

I thought the best ways was to assume values for b in the 2nd equation

Let b = -5; The equation becomes 2x^2-5x+2=0

Roots are 1/2 or 2.

Roots of 1st equation are 3/2 or 3

try to form the 1st equation

(x-3/2) (x-3) = 0

Make sure that the coefficients of x match in both the equation. (as they are equal to b).

Only one of the answer choices meet the criteria ... May I reserve the answer at this stage ...

Look forward to know better methods
thanks........can u explain me little bit clear......with specific formulas......
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by sheelanadh » Wed Aug 11, 2010 4:03 am
4GMAT_Mumbai wrote:Hi, Interesting q ... Thanks.

I thought the best ways was to assume values for b in the 2nd equation

Let b = -5; The equation becomes 2x^2-5x+2=0

Roots are 1/2 or 2.

Roots of 1st equation are 3/2 or 3

try to form the 1st equation

(x-3/2) (x-3) = 0

Make sure that the coefficients of x match in both the equation. (as they are equal to b).

Only one of the answer choices meet the criteria ... May I reserve the answer at this stage ...

Look forward to know better methods
thank you very much...
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