If #2+sqrt3# and #2-sqrt3# are solutions of #2x^2+px+q=0#, then what is #p+q#?

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Answer 1 · 2016-12-09T20:56:16

#-2#

We know that

#(x+r_1)(x+r_2) = x^2+(r_1+r_2)x+r_1r_2# then

#2(x+r_1)(x+r_2) = 2x^2+2(r_1+r_2)x+2r_1r_2#

Calling #r_1 =-( 2-sqrt(3))# and #r_2 = -(2+sqrt(3))# and comparing with

#2x^2+px+q# we have

#p=2(r_1+r_2) = -2 xx 2 = -4# and

#q = 2r_1r_2 = 2(2^2-3)=2# then #p+q=-2#