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 $p+q=-2$

If 2+sqrt32+sqrt3 and 2-sqrt32-sqrt3 are solutions of 2x^2+px+q=02x^2+px+q=0, then what is p+qp+q?