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?

Feb 25, 2017

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

Explanation:

Given:

${x}^{2} - 7 x + 12 = 0$

Divide through by $12 {x}^{2}$ to get:

$\frac{1}{12} - \frac{7}{12} \left(\frac{1}{x}\right) + {\left(\frac{1}{x}\right)}^{2} = 0$

So putting $y = \frac{1}{x}$ and transposing, we get:

${y}^{2} - \frac{7}{12} y + \frac{1}{12} = 0$

So $b = - \frac{7}{12}$ and $c = \frac{1}{12}$

$b + c = - \frac{7}{12} + \frac{1}{12} = - \frac{6}{12} = - \frac{1}{2}$