ax^2 +bx+c=0 a,b,c and two roots of the eqn. are 5 consecutive integers in some order. Find their values.


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Question Number 212354 by RojaTaniya last updated on 11/Oct/24

${a x}^{2} + b x + c = 0$ $a, b, c a n d t w o r o o t s o f t h e e q n .$ $a r e 5 c o n s e c u t i v e i n t e g e r s i n$ $s o m e o r d e r . F i n d t h e i r v a l u e s .$
	Answered by Rasheed.Sindhi last updated on 12/Oct/24

	$A Try . . .$ $∙ l e t α & β a r e r o o t s o f {a x}^{2} + b x + c = 0$ $∙ a, b, c, α, β \in Z$ $∙ α + β = \frac{- b}{a} \in Z, α β = \frac{c}{a} \in Z$ $\Rightarrow a d i v i d e s b & c$ $∙ l e t b = m a, c = n a$ $α, β = \frac{- m a \pm \sqrt{m^{2} a^{2} - 4 a (n a)}}{2 a}$ $= \frac{- m \pm \sqrt{m^{2} - 4 n}}{2}$ $m^{2} - 4 n i s p e r f e c t s q u a r e$ $a n d$ $\frac{- m \pm \sqrt{m^{2} - 4 n}}{2} \in Z$ $\underset{―}{m^{2} - 4 n i s p e r f e c t s q u a r e}$ $m^{2} = 0, 1, 4, 9, . . .$ $m^{2} = 0 \Rightarrow n = 0, - 1, - 4,$ $m^{2} = 1 \Rightarrow n = 0, - 2,$ $m^{2} = 4 \Rightarrow n = 0, 1, - 3,$ $\begin{array}{cccccc} m & n & m^{2} - 4 n & \frac{- m \pm \sqrt{m^{2} - 4 n}}{2} \\ 0 & 0, - 1, - 4, . . . & 0, 4, 16, . . . \\ 1 & 0, - 2, \\ - 1 & 0, - 2 \\ 2 & 0, 1, - 3, \\ - 2 & 0, 1, - 3, \end{array}$
	Commented by Frix last updated on 12/Oct/24
	$This narrows the possibilities: Let L={a, b, c, x_1 , x_2 } 1≤max L −min L ≤4 I think there′s only one solution...$
	$This narrows the possibilities :$ $Let L = {a, b, c, x_{1}, x_{2}}$ $1 ⩽ \max L - \min L ⩽ 4$ $I think {there}^{'} s only one solution . . .$
	Commented by Rasheed.Sindhi last updated on 13/Oct/24

	$T han k s sir!$
	Answered by Frix last updated on 12/Oct/24
	$a(x−p)(x−q)=0 ax^2 −a(p+q)x+apq=0 a, p, q, −a(p+q), apq are distinct⇒ a≠0∧p≠0∧q≠0 Let q=p+k; k∈{1, 2, 3, 4}; p+k≠0 ax^2 −a(2p+k)x+ap(p+k)=0 LetL_k ={a, p, p+k, −a(2p+k), ap(p+k)} ∣a+a(2p+k)∣=∣a(2p+k+1)∣≤4 ∣a−ap(p+k)∣=∣a(p^2 +pk−1)∣≤4 ⇒ ∣a∣≤4∧∣2p+k+1∣≤4∧∣p^2 +pk−1∣≤4 ∣a∣≤4 ⇒ a=±1, ±2, ±3, ±4 ∣2p+k+1∣≤4∧∣p^2 +pk−1∣≤4 ⇒ [remember p≠0∧p+k≠0] k=1 ⇒ p=−2, 1 k=2 ⇒ p=−3, −1 k=3 ⇒ p=−4, −2, −1 k=4 ⇒ p=−3, −1 L_1 ={a, p, p+1, −a(2p+1), ap(p+1)} p=−2 L_1 ={a, −2, −1, 3a, 2a} impossible p=1 L_1 ={a, 1, 2, −3a, 2a} impossible L_2 ={a, p, p+2, −2a(p+1), ap(p+2)} p=−3 L_2 ={a, −3, −1, 4a, 3a} impossible p=−1 L_2 ={a, −1, 1, 0, −a} a=±2 ★ L_3 ={a, p, p+3, −a(2p+3), ap(p+3)} p=−4 L_3 ={a, −4, −1, 5a, 4a} impossible p=−2 L_3 ={a, −2, 1, a, −2a} impossible p=−1 L_3 ={a, −1, 2, −a, −2a} impossible L_4 ={a, p, p+4, −2a(p+2), ap(p+4)} p=−3 L_4 ={a, −3, 1, 2a, −3a} impossible p=−1 L_4 ={a, −1, 3, −2a, −3a} impossible ★ We get 2 solutions 1. a=−2∧b=0∧c=2∧p=−1∧q=1 −2x^2 +2=0 2. a=2∧b=0∧c=−2∧p=−1∧q=1 2x^2 −2=0$
	$a (x - p) (x - q) = 0$ ${a x}^{2} - a (p + q) x + a p q = 0$ $a, p, q, - a (p + q), a p q are distinct \Rightarrow$ $a \neq 0 \land p \neq 0 \land q \neq 0$ $Let q = p + k; k \in {1, 2, 3, 4}; p + k \neq 0$ ${a x}^{2} - a (2 p + k) x + a p (p + k) = 0$ $Let L_{k} = {a, p, p + k, - a (2 p + k), a p (p + k)}$ $∣ a + a (2 p + k) ∣=∣ a (2 p + k + 1) ∣⩽ 4$ $∣ a - a p (p + k) ∣=∣ a (p^{2} + p k - 1) ∣⩽ 4$ $\Rightarrow ∣ a ∣⩽ 4 \land ∣ 2 p + k + 1 ∣⩽ 4 \land ∣ p^{2} + p k - 1 ∣⩽ 4$ $∣ a ∣⩽ 4 \Rightarrow a = \pm 1, \pm 2, \pm 3, \pm 4$ $∣ 2 p + k + 1 ∣⩽ 4 \land ∣ p^{2} + p k - 1 ∣⩽ 4 \Rightarrow$ $[remember p \neq 0 \land p + k \neq 0]$ $k = 1 \Rightarrow p = - 2, 1$ $k = 2 \Rightarrow p = - 3, - 1$ $k = 3 \Rightarrow p = - 4, - 2, - 1$ $k = 4 \Rightarrow p = - 3, - 1$ $L_{1} = {a, p, p + 1, - a (2 p + 1), a p (p + 1)}$ $p = - 2$ $L_{1} = {a, - 2, - 1, 3 a, 2 a} impossible$ $p = 1$ $L_{1} = {a, 1, 2, - 3 a, 2 a} impossible$ $L_{2} = {a, p, p + 2, - 2 a (p + 1), a p (p + 2)}$ $p = - 3$ $L_{2} = {a, - 3, - 1, 4 a, 3 a} impossible$ $p = - 1$ $L_{2} = {a, - 1, 1, 0, - a} a = \pm 2 ★$ $L_{3} = {a, p, p + 3, - a (2 p + 3), a p (p + 3)}$ $p = - 4$ $L_{3} = {a, - 4, - 1, 5 a, 4 a} impossible$ $p = - 2$ $L_{3} = {a, - 2, 1, a, - 2 a} impossible$ $p = - 1$ $L_{3} = {a, - 1, 2, - a, - 2 a} impossible$ $L_{4} = {a, p, p + 4, - 2 a (p + 2), a p (p + 4)}$ $p = - 3$ $L_{4} = {a, - 3, 1, 2 a, - 3 a} impossible$ $p = - 1$ $L_{4} = {a, - 1, 3, - 2 a, - 3 a} impossible$ $★$ $We get 2 solutions$ $1 . a = - 2 \land b = 0 \land c = 2 \land p = - 1 \land q = 1$ $- 2 x^{2} + 2 = 0$ $2 . a = 2 \land b = 0 \land c = - 2 \land p = - 1 \land q = 1$ $2 x^{2} - 2 = 0$
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