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Question-88245




Question Number 88245 by jagoll last updated on 09/Apr/20
Answered by john santu last updated on 09/Apr/20
⇒(Δ/4) = 0 , [ Δ = discriminant ]  (ab+bc)−(a^2 +b^2 )(b^2 +c^2 ) = 0  (a^2 b^2 +2ab^2 c+b^2 c^2 )−(a^2 b^2 +a^2 c+b^4 +b^2 c^2 ) =0  2ab^2 c−a^2 c^2 −b^4  = 0  (b^2 −ac)^2  = 0 ⇒ b^2  = ac   r = (b/a) = (c/b)  that prove these are in GP  r = x. let r = (b/a) = (c/b)  (a^2 +b^2 )r^2  − 2b(a+c)r + b^2 +c^2   = (a^2 +b^2 )((c/b))^2 −2b(a+c)((c/b))+  (b^2 +c^2 )   = (((ac)^2 )/b^2 ) + c^2 −2ac−c^2 +b^2   we know ac = b^2  , let substitute  = (b^4 /b^2 ) +c^2 −2b^2 −c^2 +b^2   = b^2 −b^2  = 0 ⋮
$$\Rightarrow\frac{\Delta}{\mathrm{4}}\:=\:\mathrm{0}\:,\:\left[\:\Delta\:=\:{discriminant}\:\right] \\ $$$$\left({ab}+{bc}\right)−\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\:=\:\mathrm{0} \\ $$$$\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +\mathrm{2}{ab}^{\mathrm{2}} {c}+{b}^{\mathrm{2}} {c}^{\mathrm{2}} \right)−\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}+{b}^{\mathrm{4}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} \right)\:=\mathrm{0} \\ $$$$\mathrm{2}{ab}^{\mathrm{2}} {c}−{a}^{\mathrm{2}} {c}^{\mathrm{2}} −{b}^{\mathrm{4}} \:=\:\mathrm{0} \\ $$$$\left({b}^{\mathrm{2}} −{ac}\right)^{\mathrm{2}} \:=\:\mathrm{0}\:\Rightarrow\:{b}^{\mathrm{2}} \:=\:{ac}\: \\ $$$${r}\:=\:\frac{{b}}{{a}}\:=\:\frac{{c}}{{b}} \\ $$$${that}\:{prove}\:{these}\:{are}\:{in}\:{GP} \\ $$$${r}\:=\:{x}.\:{let}\:{r}\:=\:\frac{{b}}{{a}}\:=\:\frac{{c}}{{b}} \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right){r}^{\mathrm{2}} \:−\:\mathrm{2}{b}\left({a}+{c}\right){r}\:+\:{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$$=\:\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left(\frac{{c}}{{b}}\right)^{\mathrm{2}} −\mathrm{2}{b}\left({a}+{c}\right)\left(\frac{{c}}{{b}}\right)+ \\ $$$$\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\: \\ $$$$=\:\frac{\left({ac}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:+\:{c}^{\mathrm{2}} −\mathrm{2}{ac}−{c}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$${we}\:{know}\:{ac}\:=\:{b}^{\mathrm{2}} \:,\:{let}\:{substitute} \\ $$$$=\:\frac{{b}^{\mathrm{4}} }{{b}^{\mathrm{2}} }\:+{c}^{\mathrm{2}} −\mathrm{2}{b}^{\mathrm{2}} −{c}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$$=\:{b}^{\mathrm{2}} −{b}^{\mathrm{2}} \:=\:\mathrm{0}\:\vdots \\ $$$$ \\ $$
Commented by jagoll last updated on 09/Apr/20
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by $@ty@m123 last updated on 09/Apr/20
{2b(c+a)}^2 =4(a^2 +b^2 )(b^2 +c^2 )  b^2 (c^2 +a^2 +2ca)=a^2 b^2 +a^2 c^2 +b^4 +b^2 c^2   2b^2 ac=b^4 +a^2 c^2   b^2 +a^2 c^2 −2b^2 ac=0  (b^2 −ac)^2 =0  b^2 =ac ...(1)  Now   x=−((−2b(c+a))/(2(a^2 +b^2 )))  x=((b(c+a))/(a^2 +ac))  x=((b(c+a))/(a(c+a)))  x=(b/a)  x=((bc)/b^2 ) { from(1)  x=(c/b) ... (2)  From (1) and (2)  we get the result.
$$\left\{\mathrm{2}{b}\left({c}+{a}\right)\right\}^{\mathrm{2}} =\mathrm{4}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right) \\ $$$${b}^{\mathrm{2}} \left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} +\mathrm{2}{ca}\right)={a}^{\mathrm{2}} {b}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} +{b}^{\mathrm{4}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} \\ $$$$\mathrm{2}{b}^{\mathrm{2}} {ac}={b}^{\mathrm{4}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} \\ $$$${b}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} −\mathrm{2}{b}^{\mathrm{2}} {ac}=\mathrm{0} \\ $$$$\left({b}^{\mathrm{2}} −{ac}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${b}^{\mathrm{2}} ={ac}\:…\left(\mathrm{1}\right) \\ $$$${Now}\: \\ $$$${x}=−\frac{−\mathrm{2}{b}\left({c}+{a}\right)}{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)} \\ $$$${x}=\frac{{b}\left({c}+{a}\right)}{{a}^{\mathrm{2}} +{ac}} \\ $$$${x}=\frac{{b}\left({c}+{a}\right)}{{a}\left({c}+{a}\right)} \\ $$$${x}=\frac{{b}}{{a}} \\ $$$${x}=\frac{{bc}}{{b}^{\mathrm{2}} }\:\left\{\:{from}\left(\mathrm{1}\right)\right. \\ $$$${x}=\frac{{c}}{{b}}\:…\:\left(\mathrm{2}\right) \\ $$$${From}\:\left(\mathrm{1}\right)\:{and}\:\left(\mathrm{2}\right) \\ $$$${we}\:{get}\:{the}\:{result}. \\ $$
Commented by jagoll last updated on 09/Apr/20
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$

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