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#