# The solutions of y^2+by+c=0 are the reciprocals of the solutions of x^2-7x+12=0. Find the value of b+c?

Mar 7, 2017

$b + c = - \frac{1}{2}$

#### Explanation:

$\left(x - {x}_{1}\right) \left(x - {x}_{2}\right) = {x}^{2} - \left({x}_{1} + {x}_{2}\right) x + {x}_{1} {x}_{2}$
$\left(y - \frac{1}{x} _ 1\right) \left(y - \frac{1}{x} _ 2\right) = {y}^{2} - \frac{{x}_{1} + {x}_{2}}{{x}_{1} {x}_{2}} y + \frac{1}{{x}_{1} {x}_{2}}$

so we have

$12 c = \frac{{x}_{1} {x}_{2}}{{x}_{1} {x}_{2}} = 1 \to c = \frac{1}{12}$

and ${x}_{1 , 2} = \frac{7 \pm \sqrt{49 - 48}}{2} = \left\{3 , 4\right\}$ so

$b = - \frac{3 + 4}{12} = - \frac{7}{12}$ and finally

$b + c = - \frac{1}{2}$