If $2$ is subtracted from each root,the results are reciprocals of the original roots.Find the value of $b^2+c^2+bc.$

Question

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.Find the value of $b^2+c^2+bc.$

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Let $\alpha$ and $\beta$ are the roots of the equation $x^2+bx+c=0$.
According to the question,

$\alpha-2=\frac{1}{\alpha}$ and $\beta-2=\frac{1}{\beta}$

$\alpha^2-2\alpha-1=0$ and $\beta^2-2\beta-1=0$
Adding the two equations,
$\alpha^2+\beta^2-2\alpha-2\beta-2=0$
$(\alpha+\beta)^2-2\alpha\beta-2\alpha-2\beta-2=0$

$(-b)^2-2c-2(-b)-2=0$

$b^2+2b-2c-2=0$
But i am not able to find $b^2+c^2+bc=0.$What should i do now?I am stuck here.

Answer 1

Hint...$\alpha^2-2\alpha-1=0\Rightarrow b=-2, c=-1$

Answer 2

So from equation we get quadratic which is $x^2-2x-1$ so roots if this are $1+\sqrt{2},1-\sqrt{2}$ so sum of roots is $-b/1=2,$ and product is $c/1=-1$ so $b^2+c^2+bc=4+1-2=3$ hope its clear.