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Question Number 160564 by Avijit007 last updated on 02/Dec/21
if the roots of the equation ax^2 +bx+c=0 are in the ratio 3:4,then show that 12b^2 =49ac.
$${if}\:{the}\:{roots}\:{of}\:{th}\mathrm{e}\:{equation}\:\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}=\mathrm{0} \\ $$$${are}\:{in}\:{the}\:{ratio}\:\mathrm{3}:\mathrm{4},{then}\:{show}\:{that}\: \\ $$$$\mathrm{12b}^{\mathrm{2}} =\mathrm{49ac}. \\ $$
Commented by otchereabdullai@gmail.com last updated on 02/Dec/21
nice!
$$\mathrm{nice}! \\ $$
Commented by cortano last updated on 02/Dec/21
ax^2 +bx+c=0 { (x_1 ),(x_2 ) :} →(x_1 /x_2 ) = (3/4) { ((x_1 =3k)),((x_2 =4k)) :} → { ((x_1 +x_2 =−(b/a)⇒k=−(b/(7a)))),((x_1 .x_2 =(c/a)⇒c=12ak^2 )) :} ⇒c=12a(−(b/(7a)))^2 ⇒49a^2 c=12ab ⇒49ac=12b
$$\:\mathrm{ax}^{\mathrm{2}} +\mathrm{bx}+\mathrm{c}=\mathrm{0\begin{cases}{\mathrm{x}_{\mathrm{1}} }\\{\mathrm{x}_{\mathrm{2}} }\end{cases}}\:\rightarrow\frac{\mathrm{x}_{\mathrm{1}} }{\mathrm{x}_{\mathrm{2}} }\:=\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\:\begin{cases}{\mathrm{x}_{\mathrm{1}} =\mathrm{3k}}\\{\mathrm{x}_{\mathrm{2}} =\mathrm{4k}}\end{cases}\:\rightarrow\begin{cases}{\mathrm{x}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} =−\frac{\mathrm{b}}{\mathrm{a}}\Rightarrow\mathrm{k}=−\frac{\mathrm{b}}{\mathrm{7a}}}\\{\mathrm{x}_{\mathrm{1}} .\mathrm{x}_{\mathrm{2}} =\frac{\mathrm{c}}{\mathrm{a}}\Rightarrow\mathrm{c}=\mathrm{12ak}^{\mathrm{2}} }\end{cases} \\ $$$$\Rightarrow\mathrm{c}=\mathrm{12a}\left(−\frac{\mathrm{b}}{\mathrm{7a}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{49a}^{\mathrm{2}} \mathrm{c}=\mathrm{12ab}\:\Rightarrow\mathrm{49ac}=\mathrm{12b} \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 02/Dec/21
((−b+(√(b^2 −4ac)) )/(2a)):((−b−(√(b^2 −4ac)) )/(2a))=4:3 3(((−b+(√(b^2 −4ac)) )/(2a)))=4(((−b−(√(b^2 −4ac)) )/(2a))) 4(((−b−(√(b^2 −4ac)) )/(2a)))−3(((−b+(√(b^2 −4ac)) )/(2a)))=0 −4b−4(√(b^2 −4ac)) +3b−3(√(b^2 −4ac)) −b=7(√(b^2 −4ac)) b^2 =49b^2 −196ac 48b^2 =196ac 12b^2 =49ac
$$\frac{−{b}+\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\:}{\mathrm{2}{a}}:\frac{−{b}−\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\:}{\mathrm{2}{a}}=\mathrm{4}:\mathrm{3} \\ $$$$\mathrm{3}\left(\frac{−{b}+\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\:}{\mathrm{2}{a}}\right)=\mathrm{4}\left(\frac{−{b}−\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\:}{\mathrm{2}{a}}\right) \\ $$$$\mathrm{4}\left(\frac{−{b}−\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\:}{\mathrm{2}{a}}\right)−\mathrm{3}\left(\frac{−{b}+\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\:}{\mathrm{2}{a}}\right)=\mathrm{0} \\ $$$$−\mathrm{4}{b}−\mathrm{4}\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\:+\mathrm{3}{b}−\mathrm{3}\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}} \\ $$$$−{b}=\mathrm{7}\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\: \\ $$$${b}^{\mathrm{2}} =\mathrm{49}{b}^{\mathrm{2}} −\mathrm{196}{ac} \\ $$$$\mathrm{48}{b}^{\mathrm{2}} =\mathrm{196}{ac} \\ $$$$\mathrm{12}{b}^{\mathrm{2}} =\mathrm{49}{ac} \\ $$
Commented by Rasheed.Sindhi last updated on 02/Dec/21
((−b+(√(b^2 −4ac)) )/(2a)):((−b−(√(b^2 −4ac)) )/(2a))=3:4 4(((−b+(√(b^2 −4ac)) )/(2a)))=3(((−b−(√(b^2 −4ac)) )/(2a))) −4b+4(√(b^2 −4ac)) =−3b−3(√(b^2 −4ac)) b=7(√(b^2 −4ac)) b^2 =49(b^2 −4ac)=49b^2 −196ac 48b^2 =196ac 12b^2 =49ac
$$\frac{−{b}+\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\:}{\mathrm{2}{a}}:\frac{−{b}−\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\:}{\mathrm{2}{a}}=\mathrm{3}:\mathrm{4} \\ $$$$\mathrm{4}\left(\frac{−{b}+\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\:}{\mathrm{2}{a}}\right)=\mathrm{3}\left(\frac{−{b}−\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\:}{\mathrm{2}{a}}\right) \\ $$$$−\mathrm{4}{b}+\mathrm{4}\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}}\:=−\mathrm{3}{b}−\mathrm{3}\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}} \\ $$$${b}=\mathrm{7}\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{ac}} \\ $$$${b}^{\mathrm{2}} =\mathrm{49}\left({b}^{\mathrm{2}} −\mathrm{4}{ac}\right)=\mathrm{49}{b}^{\mathrm{2}} −\mathrm{196}{ac} \\ $$$$\mathrm{48}{b}^{\mathrm{2}} =\mathrm{196}{ac} \\ $$$$\mathrm{12}{b}^{\mathrm{2}} =\mathrm{49}{ac} \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 02/Dec/21
Let the roots are 3α & 4α 3α+4α=−(b/a) ∧ 3α.4α=(c/a) 7α=−(b/a) ∧ 12α^2 =(c/a) ⇒12(−(b/(7a)))^2 =(c/a)⇒((12b^2 )/(49a^2 ))=(c/a) ⇒12b^2 =49ac
$${Let}\:{the}\:{roots}\:{are}\:\mathrm{3}\alpha\:\&\:\mathrm{4}\alpha \\ $$$$\mathrm{3}\alpha+\mathrm{4}\alpha=−\frac{{b}}{{a}}\:\wedge\:\mathrm{3}\alpha.\mathrm{4}\alpha=\frac{{c}}{{a}} \\ $$$$\mathrm{7}\alpha=−\frac{{b}}{{a}}\:\wedge\:\mathrm{12}\alpha^{\mathrm{2}} =\frac{{c}}{{a}} \\ $$$$\Rightarrow\mathrm{12}\left(−\frac{{b}}{\mathrm{7}{a}}\right)^{\mathrm{2}} =\frac{{c}}{{a}}\Rightarrow\frac{\mathrm{12}{b}^{\mathrm{2}} }{\mathrm{49}{a}^{\mathrm{2}} }=\frac{{c}}{{a}} \\ $$$$\Rightarrow\mathrm{12}{b}^{\mathrm{2}} =\mathrm{49}{ac} \\ $$$$ \\ $$
Commented by Avijit007 last updated on 02/Dec/21
thanks.
$${thanks}. \\ $$
Answered by Rasheed.Sindhi last updated on 02/Dec/21
Roots: 3k,4k { ((a(3k)^2 +b(3k)+c=0)),((a(4k)^2 +b(4k)+c=0)) :} { ((9ak^2 +3bk+c=0.......(i))),((16ak^2 +4bk+c=0.......(ii))) :} (ii)−(i):7ak^2 +bk=0 k(7ak+b)=0 7ak+b=0 k=−(b/(7a)) ;k≠0 Update Roots:3(−(b/(7a))) & 4(−(b/(7a))) Product of roots: 12(−(b/(7a)))^2 =(c/a) 12b^2 =49ac
$${Roots}:\:\mathrm{3}{k},\mathrm{4}{k} \\ $$$$\:\begin{cases}{{a}\left(\mathrm{3}{k}\right)^{\mathrm{2}} +{b}\left(\mathrm{3}{k}\right)+{c}=\mathrm{0}}\\{{a}\left(\mathrm{4}{k}\right)^{\mathrm{2}} +{b}\left(\mathrm{4}{k}\right)+{c}=\mathrm{0}}\end{cases}\: \\ $$$$\:\begin{cases}{\mathrm{9}{ak}^{\mathrm{2}} +\mathrm{3}{bk}+{c}=\mathrm{0}…….\left({i}\right)}\\{\mathrm{16}{ak}^{\mathrm{2}} +\mathrm{4}{bk}+{c}=\mathrm{0}…….\left({ii}\right)}\end{cases}\: \\ $$$$\left({ii}\right)−\left({i}\right):\mathrm{7}{ak}^{\mathrm{2}} +{bk}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:{k}\left(\mathrm{7}{ak}+{b}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{7}{ak}+{b}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:{k}=−\frac{{b}}{\mathrm{7}{a}}\:\:\:\:\:;{k}\neq\mathrm{0} \\ $$$${Update}\:{Roots}:\mathrm{3}\left(−\frac{{b}}{\mathrm{7}{a}}\right)\:\:\&\:\:\mathrm{4}\left(−\frac{{b}}{\mathrm{7}{a}}\right) \\ $$$${Product}\:{of}\:{roots}: \\ $$$$\:\:\:\:\:\:\:\:\mathrm{12}\left(−\frac{{b}}{\mathrm{7}{a}}\right)^{\mathrm{2}} =\frac{{c}}{{a}} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{12}{b}^{\mathrm{2}} =\mathrm{49}{ac} \\ $$

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