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Question Number 202497 by MATHEMATICSAM last updated on 28/Dec/23

If the difference of the roots of x^2 + 2px + q = 0 is equal to the difference of the roots of x^2 + 2qx + p = 0 [p ≠ q] then show that p + q + 1 = 0.

Ifthedifferenceoftherootsofx2+2px+q=0isequaltothedifferenceoftherootsofx2+2qx+p=0[p≠q]thenshowthatp+q+1=0.

Answered by BaliramKumar last updated on 28/Dec/23

α+β = −2p, αβ = q α^′ +β′ = −2q, α^′ β^′ = p α−β = α′−β′ (α−β)^2 = (α′−β′)^2 (α+β)^2 −4αβ = (α′+β′)^2 −4α^′ β′ (−2p)^2 −4q = (−2q)^2 −4p 4p^2 −4q = 4q^2 −4p p^2 −q = q^2 −p p^2 −q^2 +p−q=0 (p+q)(p−q)+(p−q)=0 (p−q)(p+q+1)=0 p+q+1=0 p−q≠0

α+β=−2p,αβ=qα′+β′=−2q,α′β′=pα−β=α′−β′(α−β)2=(α′−β′)2(α+β)2−4αβ=(α′+β′)2−4α′β′(−2p)2−4q=(−2q)2−4p4p2−4q=4q2−4pp2−q=q2−pp2−q2+p−q=0(p+q)(p−q)+(p−q)=0(p−q)(p+q+1)=0p+q+1=0p−q≠0

Answered by esmaeil last updated on 28/Dec/23

r_1 −r_2 =(√(p^2 −q))=r_3 −r_4 =(√(q^2 −p))= (((√δ)/(∣a∣)))→ p^2 −q^2 +p−q=0→(p−q)(p+q+1)=0

r1−r2=p2−q=r3−r4=q2−p=(δ∣a∣)→p2−q2+p−q=0→(p−q)(p+q+1)=0

Answered by MM42 last updated on 28/Dec/23

(((√Δ)/a))^2 =(((√(Δ′))/(a′)))^2 ⇒4p^2 −4q=4q^2 −4p ⇒(p−q)(p−q+1)=0 ⇒^(p≠q) p+q+1=0 ✓

(Δa)2=(Δ′a′)2⇒4p2−4q=4q2−4p⇒(p−q)(p−q+1)=0⇒p≠qp+q+1=0✓

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