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?

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Answer 1 · 2017-02-25T15:39:33

$b+c = -1/2$

Given:

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

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

$1/12-7/12(1/x)+(1/x)^2 = 0$

So putting $y = 1/x$ and transposing, we get:

$y^2-7/12y+1/12 = 0$

So $b = -7/12$ and $c = 1/12$

$b+c = -7/12+1/12 = -6/12 = -1/2$

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