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The equation x^2+bx+c=0 has distinct roots.if 2 is subtracted from each root, the results are reciprocals of the original roots. then the value of (b^2+c^2+ bc) ?

The equation x^2+bx+c=0 has distinct roots.if 2 is subtracted from each root, the results are reciprocals of the original roots. then the value of (b^2+c^2+ bc) ?

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3 Answers

Jayesh Muley
20 Points
11 years ago

let the roots be a and b.

now subtracting 2 from each root ..

a-2 &b-2 r equal to reciprocal of original roots i.e a&b .

so,a-2=1/a &b-2=1/2

solving we get a=& b=1±√2

so we take a=1+√2 & b=1-√2

so

b^2+c^2+bc=(a+b)2 +ab2 +(a+b)*ab=4+1-2=3

hence the answer is 3

Hi bbcccvb
14 Points
6 years ago
7..above answer provided is wrong... right answer is 7........nhgdsdhhjkhfddsaaazxcvvhjjjjjjgfdssd. Bnhfdssdfhjihgcxxdsg
Shrish pandey
13 Points
5 years ago
 

let the roots be a and b.

now subtracting 2 from each root ..

a-2 &b-2 r equal to reciprocal of original roots i.e a&b .

so,a-2=1/a &b-2=1/2

solving we get a=& b=1±√2

so we take a=1+√2 & b=1-√2

so

 

b^2+c^2+bc=(-a-b)2 +ab2 +(-a-b)*ab=4+1+2=7

hence the answer is 7

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