ax-2-bx-c-0-a-b-c-and-two-roots-of-the-eqn-are-5-consecutive-integers-in-some-order-Find-their-values-

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 \dots

∙ 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, \dots

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, \dots & 0, 4, 16, \dots \\ 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 \dots

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

{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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