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Question Number 26680 by Sr@2004 last updated on 28/Dec/17
(b−c)x^2 +(c−a)x+(a−b)=0 if the   eqation roots  are eqal you proved that  2b=a+c.
$$\left({b}−{c}\right){x}^{\mathrm{2}} +\left({c}−{a}\right){x}+\left({a}−{b}\right)=\mathrm{0}\:{if}\:{the}\: \\ $$$${eqation}\:{roots}\:\:{are}\:{eqal}\:{you}\:{proved}\:{that} \\ $$$$\mathrm{2}{b}={a}+{c}. \\ $$
Answered by mrW1 last updated on 28/Dec/17
equal roots ⇒ Δ=0  Δ=(c−a)^2 −4(b−c)(a−b)=0  ⇒c^2 −2ac+a^2 −4(ab−ac−b^2 +bc)=0  ⇒c^2 +a^2 −4ab+2ac+4b^2 −4bc=0  ⇒(a+c)^2 −4(a+c)b+4b^2 =0  ⇒(a+c−2b)^2 =0  ⇒2b=a+c
$${equal}\:{roots}\:\Rightarrow\:\Delta=\mathrm{0} \\ $$$$\Delta=\left({c}−{a}\right)^{\mathrm{2}} −\mathrm{4}\left({b}−{c}\right)\left({a}−{b}\right)=\mathrm{0} \\ $$$$\Rightarrow{c}^{\mathrm{2}} −\mathrm{2}{ac}+{a}^{\mathrm{2}} −\mathrm{4}\left({ab}−{ac}−{b}^{\mathrm{2}} +{bc}\right)=\mathrm{0} \\ $$$$\Rightarrow{c}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{4}{ab}+\mathrm{2}{ac}+\mathrm{4}{b}^{\mathrm{2}} −\mathrm{4}{bc}=\mathrm{0} \\ $$$$\Rightarrow\left({a}+{c}\right)^{\mathrm{2}} −\mathrm{4}\left({a}+{c}\right){b}+\mathrm{4}{b}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\left({a}+{c}−\mathrm{2}{b}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}{b}={a}+{c} \\ $$

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