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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) ?

```
3 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
```
3 years ago
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