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Question Number 103620 by bobhans last updated on 16/Jul/20

If a^2 −bc, b^2 −ac, c^2 −ab is AP where a+c = 12, find the value of a+b+c

$${If}\:{a}^{\mathrm{2}} −{bc},\:{b}^{\mathrm{2}} −{ac},\:{c}^{\mathrm{2}} −{ab}\:{is}\:{AP}\:{where}\:{a}+{c} \\ $$$$=\:\mathrm{12},\:{find}\:{the}\:{value}\:{of}\:{a}+{b}+{c}\: \\ $$

Answered by bemath last updated on 16/Jul/20

⇒AP : 2(b^2 −ac)=a^2 −bc+c^2 −ab 2b^2 −2ac = a^2 +c^2 −ab−bc 2b^2 −2ac = (a+c)^2 −2ac−ab−bc 2b^2 = 144−ab−b(12−a) 2b^2 = 144−12b b^2 +6b−72 = 0 ⇒(b+12)(b−6)=0→ { ((b=6)),((b=−12)) :} ∴ a+b+c = 18 or zero

$$\Rightarrow{AP}\::\:\mathrm{2}\left({b}^{\mathrm{2}} −{ac}\right)={a}^{\mathrm{2}} −{bc}+{c}^{\mathrm{2}} −{ab} \\ $$$$\mathrm{2}{b}^{\mathrm{2}} −\mathrm{2}{ac}\:=\:{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc} \\ $$$$\mathrm{2}{b}^{\mathrm{2}} −\mathrm{2}{ac}\:=\:\left({a}+{c}\right)^{\mathrm{2}} −\mathrm{2}{ac}−{ab}−{bc} \\ $$$$\mathrm{2}{b}^{\mathrm{2}} \:=\:\mathrm{144}−{ab}−{b}\left(\mathrm{12}−{a}\right) \\ $$$$\mathrm{2}{b}^{\mathrm{2}} \:=\:\mathrm{144}−\mathrm{12}{b} \\ $$$${b}^{\mathrm{2}} +\mathrm{6}{b}−\mathrm{72}\:=\:\mathrm{0} \\ $$$$\Rightarrow\left({b}+\mathrm{12}\right)\left({b}−\mathrm{6}\right)=\mathrm{0}\rightarrow\begin{cases}{{b}=\mathrm{6}}\\{{b}=−\mathrm{12}}\end{cases} \\ $$$$\therefore\:{a}+{b}+{c}\:=\:\mathrm{18}\:{or}\:{zero}\: \\ $$

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