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Question Number 113684 by ZiYangLee last updated on 14/Sep/20

A nice question <3 If a quadratic equation (1−q+(p^2 /2))x^2 +p(1+q)x+q(q−1)+(p^2 /2)=0 has equal roots, prove that p^2 =4q

Anicequestion<3Ifaquadraticequation(1−q+p22)x2+p(1+q)x+q(q−1)+p22=0hasequalroots,provethatp2=4q

Commented by Dwaipayan Shikari last updated on 14/Sep/20

p^2 =4q (1−q+((4q)/2))x^2 +p(1+q)+q^2 −q+2q=0 (1+q)x^2 +p(1+q)+q(1+q)=0 It has roots α+β=−p((1+q)/(1+q))=−p αβ=q=(p^2 /4) (α−β)=(√(p^2 −4.(p^2 /4))) =0 α=β (Which is true)

p2=4q(1−q+4q2)x2+p(1+q)+q2−q+2q=0(1+q)x2+p(1+q)+q(1+q)=0Ithasrootsα+β=−p1+q1+q=−pαβ=q=p24(α−β)=p2−4.p24=0α=β(Whichistrue)

Answered by MJS_new last updated on 14/Sep/20

(1−q+(p^2 /2))x^2 +p(1+q)x+q(q−1)+(p^2 /2)=0 x^2 +((2p(q+1))/(p^2 −2(q−1)))x+((p^2 +2q(q−1))/(p^2 −2(q−1)))=0 x=t−((p(q+1))/(p^2 −2(q−1))) t^2 −(((p^2 −4q)(p^2 +q^2 −2q+1))/((p^2 −2(q−1))^2 ))=0 x_1 =x_2 ⇔ t=0 ⇒ p^2 −4q=0∨p^2 +q^2 −2q+1=0 ⇒ p^2 =4q (p^2 +q^2 −2q+1=0 ⇒ p^2 =−(q−1)^2 ⇒ p∉R)

(1−q+p22)x2+p(1+q)x+q(q−1)+p22=0x2+2p(q+1)p2−2(q−1)x+p2+2q(q−1)p2−2(q−1)=0x=t−p(q+1)p2−2(q−1)t2−(p2−4q)(p2+q2−2q+1)(p2−2(q−1))2=0x1=x2⇔t=0⇒p2−4q=0∨p2+q2−2q+1=0⇒p2=4q(p2+q2−2q+1=0⇒p2=−(q−1)2⇒p∉R)

Commented by ZiYangLee last updated on 15/Sep/20

Cool!

Cool!

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