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Let  p and  q be real numbers such that the roots of x^2+px+q=0
are real. Prove that the roots of x^2+px+q +(x+a)(2x+p) = 0
are real, for any real number a. 
thx!

 Aug 11, 2020
edited by Mightyduck123456  Aug 11, 2020
 #1
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\(\displaystyle \text{If}\;\;x^{2}+px+q=0 \;\;\text{ has real (distinct) roots, then }\;\; p^{2}-4q>0.\)

 

\(\displaystyle x^{2}+px+q+(x+a)(2x+p)\\=x^{2}+px+q+2x^{2}+x(2a+p)+ap\\ =3x^{2}+x(2a+2p)+q+ap=0\)

 

\(\displaystyle \text{For this to have real (distinct) roots, it's necessary that}\\ (2a+2p)^{2}-12(q+ap)>0, \\ \text{and since}\\ = (2a+2p)^{2}-12(q+ap)\\=4a^{2}+4p^{2}+8ap-12q-12ap \\= 4a^{2}+p^{2}-4ap+3p^{2}-12q\\=(2a-p)^{2}+3(p^{2}-4q)\\ \text{this will be the case for any real number a.} \)

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 Aug 12, 2020

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