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Question Number 202193 by MATHEMATICSAM last updated on 22/Dec/23
If α, β are the roots of x^2 + ax + b = 0 and α + δ, β + δ are the roots of x^2 + px + q = 0 then show that a^2 − p^2 = 4(b − q).
$$\mathrm{If}\:\alpha,\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:{x}^{\mathrm{2}} \:+\:{ax}\:+\:{b}\:=\:\:\mathrm{0}\: \\ $$$$\mathrm{and}\:\alpha\:+\:\delta,\:\beta\:+\:\delta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${x}^{\mathrm{2}} \:+\:{px}\:+\:{q}\:=\:\mathrm{0}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that}\: \\ $$$${a}^{\mathrm{2}} \:−\:{p}^{\mathrm{2}} \:=\:\mathrm{4}\left({b}\:−\:{q}\right). \\ $$
Answered by Rasheed.Sindhi last updated on 22/Dec/23
{ ((x^2 + ax + b = 0 ; α+β =−a,αβ=b)),((x^2 + px + q = 0 ;^★ (α + δ)+(β + δ)=−p,^⧫ (α + δ)(β + δ)=q)) :} ^★ α+β+2δ=−p⇒−a+2δ=−p⇒a−p=2δ⇒δ=((a−p)/2) ^⧫ αβ+(α+β)δ+δ^2 =q⇒b−aδ+δ^2 ⇒b−q=aδ−δ^2 ⇒b−q=a(((a−p)/2))−(((a−p)/2))^2 ⇒b−q=(((a−p)/2))(a−((a−p)/2))=(((a−p)/2))(((a+p)/2)) ⇒4(b−q)=a^2 −p^2 QED
$$\begin{cases}{{x}^{\mathrm{2}} \:+\:{ax}\:+\:{b}\:=\:\:\mathrm{0}\:;\:\alpha+\beta\:=−{a},\alpha\beta={b}}\\{{x}^{\mathrm{2}} \:+\:{px}\:+\:{q}\:=\:\mathrm{0}\:;\:^{\bigstar} \left(\alpha\:+\:\delta\right)+\left(\beta\:+\:\delta\right)=−{p},\:^{\blacklozenge} \left(\alpha\:+\:\delta\right)\left(\beta\:+\:\delta\right)={q}}\end{cases} \\ $$$$\:^{\bigstar} \:\alpha+\beta+\mathrm{2}\delta=−{p}\Rightarrow−{a}+\mathrm{2}\delta=−{p}\Rightarrow{a}−{p}=\mathrm{2}\delta\Rightarrow\delta=\frac{{a}−{p}}{\mathrm{2}} \\ $$$$\:^{\blacklozenge} \:\alpha\beta+\left(\alpha+\beta\right)\delta+\delta^{\mathrm{2}} ={q}\Rightarrow{b}−{a}\delta+\delta^{\mathrm{2}} \Rightarrow{b}−{q}={a}\delta−\delta^{\mathrm{2}} \\ $$$$\:\:\Rightarrow{b}−{q}={a}\left(\frac{{a}−{p}}{\mathrm{2}}\right)−\left(\frac{{a}−{p}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\:\Rightarrow{b}−{q}=\left(\frac{{a}−{p}}{\mathrm{2}}\right)\left({a}−\frac{{a}−{p}}{\mathrm{2}}\right)=\left(\frac{{a}−{p}}{\mathrm{2}}\right)\left(\frac{{a}+{p}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\mathrm{4}\left({b}−{q}\right)={a}^{\mathrm{2}} −{p}^{\mathrm{2}} \\ $$$$\:\:\:\:\mathrm{QED} \\ $$
Answered by MM42 last updated on 22/Dec/23
(α+δ)+(β+δ)=−p⇒δ=((a−p)/2) (α+δ)(β+δ)=αβ+(α+β)δ+δ^2 q=b−aδ+δ^2 =b−a(((a−p)/2))+(((a−p)/2))^2 4q=4b−2a^2 +2ap+a^2 −2ap+p^2 ⇒a^2 −p^2 =4(b−q) ✓
$$\left(\alpha+\delta\right)+\left(\beta+\delta\right)=−{p}\Rightarrow\delta=\frac{{a}−{p}}{\mathrm{2}} \\ $$$$\left(\alpha+\delta\right)\left(\beta+\delta\right)=\alpha\beta+\left(\alpha+\beta\right)\delta+\delta^{\mathrm{2}} \\ $$$${q}={b}−{a}\delta+\delta^{\mathrm{2}} ={b}−{a}\left(\frac{{a}−{p}}{\mathrm{2}}\right)+\left(\frac{{a}−{p}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\mathrm{4}{q}=\mathrm{4}{b}−\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{ap}+{a}^{\mathrm{2}} −\mathrm{2}{ap}+{p}^{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} −{p}^{\mathrm{2}} =\mathrm{4}\left({b}−{q}\right)\:\:\checkmark \\ $$$$ \\ $$

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