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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?

1 Answer

Mar 7, 2017

#b+c=-1/2#

Explanation:

#(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2#
#(y-1/x_1)(y-1/x_2) = y^2-(x_1+x_2)/(x_1x_2)y+1/(x_1x_2)#

so we have

#12c=(x_1x_2)/(x_1x_2)=1->c=1/12#

and #x_(1,2) = (7pm sqrt(49-48))/2={3,4}# so

#b=-(3+4)/12=-7/12# and finally

#b+c=-1/2#

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