Show-that-a-b-c-x-2-b-c-a-xy-c-a-b-y-2-will-be-a-perfect-square-if-a-b-c-are-in-H-P-

Question Number 20307 by Tinkutara last updated on 25/Aug/17

$$\mathrm{Show}\:\mathrm{that}\:{a}\left({b}\:−\:{c}\right){x}^{\mathrm{2}} \:+\:{b}\left({c}\:−\:{a}\right){xy}\:+ \\ $$$${c}\left({a}\:−\:{b}\right){y}^{\mathrm{2}} \:\mathrm{will}\:\mathrm{be}\:\mathrm{a}\:\mathrm{perfect}\:\mathrm{square}\:\mathrm{if}\:{a}, \\ $$$${b},\:{c}\:\mathrm{are}\:\mathrm{in}\:\mathrm{H}.\mathrm{P}. \\ $$

Answered by ajfour last updated on 25/Aug/17

=x^2 [c(a−b)((y/x))^2 +b(c−a)((y/x))+a(b−c)] =c(a−b)x^2 {((y/x))^2 +2[((b(c−a))/(2c(a−b)))]x+ [((b(c−a))/(2c(a−b)))]^2 }−((b^2 (c−a)^2 )/(4c(a−b)))+a(b−c) =c(a−b)x^2 {(y/x)+((b(c−a))/(2c(a−b)))}^2 + ((4ac(a−b)(b−c)−b^2 (c−a)^2 )/(4c(a−b))) ⇒ expression is a perfect square if the constant term is zero, ⇒ 4ac(ab−ac−b^2 +bc) =b^2 c^2 −2b^2 ac+a^2 b^2 ⇒ 4a^2 bc−4a^2 c^2 −4b^2 ac+4abc^2 =b^2 c^2 −2b^2 ac+a^2 b^2 ⇒ 4abc(a+c)=4a^2 c^2 +b^2 (a^2 +c^2 +2ac) ⇒ 4abc(a+c)=4a^2 c^2 +b^2 (a+c)^2 ⇒ [b(a+c)−2ac]^2 =0 ⇒ b=((2ac)/(a+c)) ⇒ a, b, c need be in H.P. •

$$={x}^{\mathrm{2}} \left[{c}\left({a}−{b}\right)\left(\frac{{y}}{{x}}\right)^{\mathrm{2}} +{b}\left({c}−{a}\right)\left(\frac{{y}}{{x}}\right)+{a}\left({b}−{c}\right)\right] \\ $$$$={c}\left({a}−{b}\right){x}^{\mathrm{2}} \left\{\left(\frac{{y}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}\left[\frac{{b}\left({c}−{a}\right)}{\mathrm{2}{c}\left({a}−{b}\right)}\right]{x}+\right. \\ $$$$\left.\:\:\:\:\left[\frac{{b}\left({c}−{a}\right)}{\mathrm{2}{c}\left({a}−{b}\right)}\right]^{\mathrm{2}} \right\}−\frac{{b}^{\mathrm{2}} \left({c}−{a}\right)^{\mathrm{2}} }{\mathrm{4}{c}\left({a}−{b}\right)}+{a}\left({b}−{c}\right) \\ $$$$={c}\left({a}−{b}\right){x}^{\mathrm{2}} \left\{\frac{{y}}{{x}}+\frac{{b}\left({c}−{a}\right)}{\mathrm{2}{c}\left({a}−{b}\right)}\right\}^{\mathrm{2}} + \\ $$$$\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{4}{ac}\left({a}−{b}\right)\left({b}−{c}\right)−{b}^{\mathrm{2}} \left({c}−{a}\right)^{\mathrm{2}} }{\mathrm{4}{c}\left({a}−{b}\right)} \\ $$$$\Rightarrow\:{expression}\:{is}\:{a}\:{perfect}\:{square} \\ $$$${if}\:{the}\:{constant}\:{term}\:{is}\:{zero}, \\ $$$$\Rightarrow\:\mathrm{4}{ac}\left({ab}−{ac}−{b}^{\mathrm{2}} +{bc}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={b}^{\mathrm{2}} {c}^{\mathrm{2}} −\mathrm{2}{b}^{\mathrm{2}} {ac}+{a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\mathrm{4}{a}^{\mathrm{2}} {bc}−\mathrm{4}{a}^{\mathrm{2}} {c}^{\mathrm{2}} −\mathrm{4}{b}^{\mathrm{2}} {ac}+\mathrm{4}{abc}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={b}^{\mathrm{2}} {c}^{\mathrm{2}} −\mathrm{2}{b}^{\mathrm{2}} {ac}+{a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{4}{abc}\left({a}+{c}\right)=\mathrm{4}{a}^{\mathrm{2}} {c}^{\mathrm{2}} +{b}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ac}\right) \\ $$$$\Rightarrow\:\mathrm{4}{abc}\left({a}+{c}\right)=\mathrm{4}{a}^{\mathrm{2}} {c}^{\mathrm{2}} +{b}^{\mathrm{2}} \left({a}+{c}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\left[{b}\left({a}+{c}\right)−\mathrm{2}{ac}\right]^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\:\:\:\:{b}=\frac{\mathrm{2}{ac}}{{a}+{c}}\: \\ $$$$\Rightarrow\:\:\:{a},\:{b},\:{c}\:{need}\:{be}\:\:{in}\:{H}.{P}.\:\bullet \\ $$

Commented by Tinkutara last updated on 25/Aug/17

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

Answered by $@ty@m last updated on 26/Aug/17

discriminant=0 ⇒{b(c−a)y}^2 −4ac(b−c)(a−b)y^2 =0 ⇒b^2 (c−a)^2 −4ac(b−c)(a−b)=0 ⇒b^2 (c−a)^2 =4ac(b−c)(a−b) ⇒{a(b−c)+c(a−b)}^2 =4ac(b−c)(a−b) ⇒{a(b−c)−c(a−b)}^2 =0 ⇒a(b−c)=c(a−b) ⇒ab−ac=ca−cb ⇒(1/c)−(1/b)=(1/b)−(1/a) (dividing by abc) ⇒a, b, c are in H.P.

$${discriminant}=\mathrm{0} \\ $$$$\Rightarrow\left\{{b}\left({c}−{a}\right){y}\right\}^{\mathrm{2}} −\mathrm{4}{ac}\left({b}−{c}\right)\left({a}−{b}\right){y}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{b}^{\mathrm{2}} \left({c}−{a}\right)^{\mathrm{2}} −\mathrm{4}{ac}\left({b}−{c}\right)\left({a}−{b}\right)=\mathrm{0} \\ $$$$\Rightarrow{b}^{\mathrm{2}} \left({c}−{a}\right)^{\mathrm{2}} =\mathrm{4}{ac}\left({b}−{c}\right)\left({a}−{b}\right) \\ $$$$\Rightarrow\left\{{a}\left({b}−{c}\right)+{c}\left({a}−{b}\right)\right\}^{\mathrm{2}} =\mathrm{4}{ac}\left({b}−{c}\right)\left({a}−{b}\right) \\ $$$$\Rightarrow\left\{{a}\left({b}−{c}\right)−{c}\left({a}−{b}\right)\right\}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{a}\left({b}−{c}\right)={c}\left({a}−{b}\right) \\ $$$$\Rightarrow{ab}−{ac}={ca}−{cb} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{c}}−\frac{\mathrm{1}}{{b}}=\frac{\mathrm{1}}{{b}}−\frac{\mathrm{1}}{{a}}\:\left({dividing}\:{by}\:{abc}\right) \\ $$$$\Rightarrow{a},\:{b},\:{c}\:{are}\:{in}\:{H}.{P}. \\ $$

Commented by $@ty@m last updated on 25/Aug/17