if the roots of the equation ax^2 +bx+c=0 are in the ratio 3:4,then show that 12b^2 =49ac.


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Question Number 160564 by Avijit007 last updated on 02/Dec/21

$i f t h e r o o t s o f t h e e q u a t i o n {ax}^{2} + bx + c = 0$ $a r e i n t h e r a t i o 3 : 4, t h e n s h o w t h a t$ ${12 b}^{2} = 49 ac .$
Commented by otchereabdullai@gmail.com last updated on 02/Dec/21

$nice!$
Commented by cortano last updated on 02/Dec/21
$ax^2 +bx+c=0 { (x_1 ),(x_2 ) :} →(x_1 /x_2 ) = (3/4) { ((x_1 =3k)),((x_2 =4k)) :} → { ((x_1 +x_2 =−(b/a)⇒k=−(b/(7a)))),((x_1 .x_2 =(c/a)⇒c=12ak^2 )) :} ⇒c=12a(−(b/(7a)))^2 ⇒49a^2 c=12ab ⇒49ac=12b$
${ax}^{2} + bx + c = 0 {\begin{cases} x_{1} \\ x_{2} \end{cases} \to \frac{x_{1}}{x_{2}} = \frac{3}{4}$ ${\begin{cases} x_{1} = 3 k \\ x_{2} = 4 k \end{cases} \to {\begin{cases} x_{1} + x_{2} = - \frac{b}{a} \Rightarrow k = - \frac{b}{7 a} \\ x_{1} . x_{2} = \frac{c}{a} \Rightarrow c = {12 ak}^{2} \end{cases}$ $\Rightarrow c = 12 a {(- \frac{b}{7 a})}^{2}$ $\Rightarrow {49 a}^{2} c = 12 ab \Rightarrow 49 ac = 12 b$
	Answered by Rasheed.Sindhi last updated on 02/Dec/21

	$\frac{- b + \sqrt{b^{2} - 4 a c}}{2 a} : \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a} = 4 : 3$ $3 (\frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}) = 4 (\frac{- b - \sqrt{b^{2} - 4 a c}}{2 a})$ $4 (\frac{- b - \sqrt{b^{2} - 4 a c}}{2 a}) - 3 (\frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}) = 0$ $- 4 b - 4 \sqrt{b^{2} - 4 a c} + 3 b - 3 \sqrt{b^{2} - 4 a c}$ $- b = 7 \sqrt{b^{2} - 4 a c}$ $b^{2} = 49 b^{2} - 196 a c$ $48 b^{2} = 196 a c$ $12 b^{2} = 49 a c$
	Commented by Rasheed.Sindhi last updated on 02/Dec/21

	$\frac{- b + \sqrt{b^{2} - 4 a c}}{2 a} : \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a} = 3 : 4$ $4 (\frac{- b + \sqrt{b^{2} - 4 a c}}{2 a}) = 3 (\frac{- b - \sqrt{b^{2} - 4 a c}}{2 a})$ $- 4 b + 4 \sqrt{b^{2} - 4 a c} = - 3 b - 3 \sqrt{b^{2} - 4 a c}$ $b = 7 \sqrt{b^{2} - 4 a c}$ $b^{2} = 49 (b^{2} - 4 a c) = 49 b^{2} - 196 a c$ $48 b^{2} = 196 a c$ $12 b^{2} = 49 a c$
	Answered by Rasheed.Sindhi last updated on 02/Dec/21

	$L e t t h e r o o t s a r e 3 α & 4 α$ $3 α + 4 α = - \frac{b}{a} \land 3 α . 4 α = \frac{c}{a}$ $7 α = - \frac{b}{a} \land 12 α^{2} = \frac{c}{a}$ $\Rightarrow 12 {(- \frac{b}{7 a})}^{2} = \frac{c}{a} \Rightarrow \frac{12 b^{2}}{49 a^{2}} = \frac{c}{a}$ $\Rightarrow 12 b^{2} = 49 a c$
	Commented by Avijit007 last updated on 02/Dec/21

	$t h a n k s .$
	Answered by Rasheed.Sindhi last updated on 02/Dec/21
	$Roots: 3k,4k { ((a(3k)^2 +b(3k)+c=0)),((a(4k)^2 +b(4k)+c=0)) :} { ((9ak^2 +3bk+c=0.......(i))),((16ak^2 +4bk+c=0.......(ii))) :} (ii)−(i):7ak^2 +bk=0 k(7ak+b)=0 7ak+b=0 k=−(b/(7a)) ;k≠0 Update Roots:3(−(b/(7a))) & 4(−(b/(7a))) Product of roots: 12(−(b/(7a)))^2 =(c/a) 12b^2 =49ac$
	$R o o t s : 3 k, 4 k$ ${\begin{cases} a {(3 k)}^{2} + b (3 k) + c = 0 \\ a {(4 k)}^{2} + b (4 k) + c = 0 \end{cases}$ ${\begin{cases} 9 {a k}^{2} + 3 b k + c = 0 . . . . . . . (i) \\ 16 {a k}^{2} + 4 b k + c = 0 . . . . . . . (i i) \end{cases}$ $(i i) - (i) : 7 {a k}^{2} + b k = 0$ $k (7 a k + b) = 0$ $7 a k + b = 0$ $k = - \frac{b}{7 a}; k \neq 0$ $U p d a t e R o o t s : 3 (- \frac{b}{7 a}) & 4 (- \frac{b}{7 a})$ $P r o d u c t o f r o o t s :$ $12 {(- \frac{b}{7 a})}^{2} = \frac{c}{a}$ $12 b^{2} = 49 a c$
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