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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-

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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.
$$\mathrm{If}\:\mathrm{the}\:\mathrm{difference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of} \\ $$$${x}^{\mathrm{2}} \:+\:\mathrm{2}{px}\:+\:{q}\:=\:\mathrm{0}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\: \\ $$$$\mathrm{difference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${x}^{\mathrm{2}} \:+\:\mathrm{2}{qx}\:+\:{p}\:=\:\mathrm{0}\:\left[{p}\:\neq\:{q}\right]\:\mathrm{then}\:\mathrm{show} \\ $$$$\mathrm{that}\:{p}\:+\:{q}\:+\:\mathrm{1}\:=\:\mathrm{0}. \\ $$
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
$$\alpha+\beta\:=\:−\mathrm{2p},\:\:\:\:\:\:\:\:\:\:\:\alpha\beta\:=\:\mathrm{q} \\ $$$$\alpha^{'} +\beta'\:=\:−\mathrm{2q},\:\:\:\:\:\:\:\:\:\:\:\alpha^{'} \beta^{'} \:=\:\mathrm{p} \\ $$$$\alpha−\beta\:=\:\alpha'−\beta' \\ $$$$\left(\alpha−\beta\right)^{\mathrm{2}} \:=\:\left(\alpha'−\beta'\right)^{\mathrm{2}} \\ $$$$\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{4}\alpha\beta\:=\:\left(\alpha'+\beta'\right)^{\mathrm{2}} −\mathrm{4}\alpha^{'} \beta' \\ $$$$\left(−\mathrm{2p}\right)^{\mathrm{2}} −\mathrm{4q}\:=\:\left(−\mathrm{2q}\right)^{\mathrm{2}} −\mathrm{4p} \\ $$$$\mathrm{4p}^{\mathrm{2}} −\mathrm{4q}\:=\:\mathrm{4q}^{\mathrm{2}} −\mathrm{4p} \\ $$$$\mathrm{p}^{\mathrm{2}} −\mathrm{q}\:=\:\mathrm{q}^{\mathrm{2}} −\mathrm{p} \\ $$$$\mathrm{p}^{\mathrm{2}} −\mathrm{q}^{\mathrm{2}} +\mathrm{p}−\mathrm{q}=\mathrm{0} \\ $$$$\left(\mathrm{p}+\mathrm{q}\right)\left(\mathrm{p}−\mathrm{q}\right)+\left(\mathrm{p}−\mathrm{q}\right)=\mathrm{0} \\ $$$$\left(\mathrm{p}−\mathrm{q}\right)\left(\mathrm{p}+\mathrm{q}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{p}+\mathrm{q}+\mathrm{1}=\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{p}−\mathrm{q}\neq\mathrm{0} \\ $$
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
$${r}_{\mathrm{1}} −{r}_{\mathrm{2}} =\sqrt{{p}^{\mathrm{2}} −{q}}={r}_{\mathrm{3}} −{r}_{\mathrm{4}} =\sqrt{{q}^{\mathrm{2}} −{p}}= \\ $$$$\left(\frac{\sqrt{\delta}}{\mid{a}\mid}\right)\rightarrow \\ $$$${p}^{\mathrm{2}} −{q}^{\mathrm{2}} +{p}−{q}=\mathrm{0}\rightarrow\left({p}−{q}\right)\left({p}+{q}+\mathrm{1}\right)=\mathrm{0} \\ $$$$ \\ $$
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 ✓
$$\left(\frac{\sqrt{\Delta}}{{a}}\right)^{\mathrm{2}} =\left(\frac{\sqrt{\Delta'}}{{a}'}\right)^{\mathrm{2}} \Rightarrow\mathrm{4}{p}^{\mathrm{2}} −\mathrm{4}{q}=\mathrm{4}{q}^{\mathrm{2}} −\mathrm{4}{p} \\ $$$$\Rightarrow\left({p}−{q}\right)\left({p}−{q}+\mathrm{1}\right)=\mathrm{0}\:\: \\ $$$$\overset{{p}\neq{q}} {\Rightarrow}\:\:{p}+{q}+\mathrm{1}=\mathrm{0}\:\:\checkmark \\ $$$$ \\ $$

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