Menu Close

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

Tinku Tara 0 Comments
Pin




Question Number 212354 by RojaTaniya last updated on 11/Oct/24
ax^2 +bx+c=0 a,b,c and two roots of the eqn. are 5 consecutive integers in some order. Find their values.
$$\:\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$$\:{a},{b},{c}\:{and}\:{two}\:{roots}\:{of}\:{the}\:{eqn}. \\ $$$$\:{are}\:\mathrm{5}\:{consecutive}\:{integers}\:{in}\: \\ $$$$\:{some}\:{order}.\:{Find}\:{their}\:{values}. \\ $$
Answered by Rasheed.Sindhi last updated on 12/Oct/24
A Try... •let α & β are roots of ax^2 +bx+c=0 •a,b,c ,α,β ∈ Z •α+β=((−b)/a) ∈ Z , αβ=(c/a) ∈ Z ⇒a divides b & c •let b=ma , c=na α,β=((−ma±(√(m^2 a^2 −4a(na))))/(2a)) =((−m±(√(m^2 −4n)))/2) m^2 −4n is perfect square and ((−m±(√(m^2 −4n)))/2) ∈ Z m^2 −4n is perfect square m^2 =0,1,4,9,... m^2 =0⇒n=0,−1,−4, m^2 =1⇒n=0,−2, m^2 =4⇒n=0,1,−3, determinant ((m,n,(m^2 −4n),( ((−m±(√(m^2 −4n)))/2))),(( 0),(0,−1,−4,...),(0,4,16,...),),(( 1),(0,−2,),,),((−1),(0,−2),,),(( 2),(0,1,−3,),,),((−2),(0,1,−3,),,))
$$\boldsymbol{\mathrm{A}}\:\boldsymbol{\mathrm{Try}}… \\ $$$$\bullet{let}\:\alpha\:\&\:\beta\:{are}\:{roots}\:{of}\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$$\bullet{a},{b},{c}\:,\alpha,\beta\:\in\:\mathbb{Z} \\ $$$$\bullet\alpha+\beta=\frac{−{b}}{{a}}\:\in\:\mathbb{Z}\:,\:\alpha\beta=\frac{{c}}{{a}}\:\in\:\mathbb{Z} \\ $$$$\:\:\:\:\Rightarrow{a}\:{divides}\:{b}\:\&\:{c} \\ $$$$\bullet{let}\:{b}={ma}\:,\:{c}={na} \\ $$$$\alpha,\beta=\frac{−{ma}\pm\sqrt{{m}^{\mathrm{2}} {a}^{\mathrm{2}} −\mathrm{4}{a}\left({na}\right)}}{\mathrm{2}{a}} \\ $$$$\:\:\:\:\:\:\:\:=\frac{−{m}\pm\sqrt{{m}^{\mathrm{2}} −\mathrm{4}{n}}}{\mathrm{2}} \\ $$$$\:\:\:\:{m}^{\mathrm{2}} −\mathrm{4}{n}\:{is}\:{perfect}\:{square} \\ $$$$\:\:\:\:{and}\: \\ $$$$\:\:\:\:\frac{−{m}\pm\sqrt{{m}^{\mathrm{2}} −\mathrm{4}{n}}}{\mathrm{2}}\:\in\:\mathbb{Z} \\ $$$$\underline{{m}^{\mathrm{2}} −\mathrm{4}{n}\:{is}\:{perfect}\:{square}} \\ $$$${m}^{\mathrm{2}} =\mathrm{0},\mathrm{1},\mathrm{4},\mathrm{9},… \\ $$$${m}^{\mathrm{2}} =\mathrm{0}\Rightarrow{n}=\mathrm{0},−\mathrm{1},−\mathrm{4}, \\ $$$${m}^{\mathrm{2}} =\mathrm{1}\Rightarrow{n}=\mathrm{0},−\mathrm{2}, \\ $$$${m}^{\mathrm{2}} =\mathrm{4}\Rightarrow{n}=\mathrm{0},\mathrm{1},−\mathrm{3}, \\ $$$$\begin{array}{|c|c|c|c|c|c|}{{m}}&\hline{{n}}&\hline{{m}^{\mathrm{2}} −\mathrm{4}{n}}&\hline{\:\frac{−{m}\pm\sqrt{{m}^{\mathrm{2}} −\mathrm{4}{n}}}{\mathrm{2}}}\\{\:\:\:\:\mathrm{0}}&\hline{\mathrm{0},−\mathrm{1},−\mathrm{4},…}&\hline{\mathrm{0},\mathrm{4},\mathrm{16},…}&\hline{}\\{\:\:\:\:\mathrm{1}}&\hline{\mathrm{0},−\mathrm{2},}&\hline{}&\hline{}\\{−\mathrm{1}}&\hline{\mathrm{0},−\mathrm{2}}&\hline{}&\hline{}\\{\:\:\:\:\mathrm{2}}&\hline{\mathrm{0},\mathrm{1},−\mathrm{3},}&\hline{}&\hline{}\\{−\mathrm{2}}&\hline{\mathrm{0},\mathrm{1},−\mathrm{3},}&\hline{}&\hline{}\\\hline\end{array} \\ $$
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...
$$\mathrm{This}\:\mathrm{narrows}\:\mathrm{the}\:\mathrm{possibilities}: \\ $$$$\mathrm{Let}\:{L}=\left\{{a},\:{b},\:{c},\:{x}_{\mathrm{1}} ,\:{x}_{\mathrm{2}} \right\} \\ $$$$\mathrm{1}\leqslant\mathrm{max}\:{L}\:−\mathrm{min}\:{L}\:\leqslant\mathrm{4} \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{there}'\mathrm{s}\:\mathrm{only}\:\mathrm{one}\:\mathrm{solution}… \\ $$
Commented by Rasheed.Sindhi last updated on 13/Oct/24
Thanks sir!
$$\mathbb{T}\boldsymbol{\mathrm{han}}\Bbbk\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{sir}}! \\ $$
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
$${a}\left({x}−{p}\right)\left({x}−{q}\right)=\mathrm{0} \\ $$$${ax}^{\mathrm{2}} −{a}\left({p}+{q}\right){x}+{apq}=\mathrm{0} \\ $$$${a},\:{p},\:{q},\:−{a}\left({p}+{q}\right),\:{apq}\:\mathrm{are}\:\mathrm{distinct}\Rightarrow \\ $$$${a}\neq\mathrm{0}\wedge{p}\neq\mathrm{0}\wedge{q}\neq\mathrm{0} \\ $$$$ \\ $$$$\mathrm{Let}\:{q}={p}+{k};\:{k}\in\left\{\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4}\right\};\:{p}+{k}\neq\mathrm{0} \\ $$$${ax}^{\mathrm{2}} −{a}\left(\mathrm{2}{p}+{k}\right){x}+{ap}\left({p}+{k}\right)=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{Let}{L}_{{k}} =\left\{{a},\:{p},\:{p}+{k},\:−{a}\left(\mathrm{2}{p}+{k}\right),\:{ap}\left({p}+{k}\right)\right\} \\ $$$$ \\ $$$$\mid{a}+{a}\left(\mathrm{2}{p}+{k}\right)\mid=\mid{a}\left(\mathrm{2}{p}+{k}+\mathrm{1}\right)\mid\leqslant\mathrm{4} \\ $$$$\mid{a}−{ap}\left({p}+{k}\right)\mid=\mid{a}\left({p}^{\mathrm{2}} +{pk}−\mathrm{1}\right)\mid\leqslant\mathrm{4} \\ $$$$\Rightarrow\:\mid{a}\mid\leqslant\mathrm{4}\wedge\mid\mathrm{2}{p}+{k}+\mathrm{1}\mid\leqslant\mathrm{4}\wedge\mid{p}^{\mathrm{2}} +{pk}−\mathrm{1}\mid\leqslant\mathrm{4} \\ $$$$ \\ $$$$\mid{a}\mid\leqslant\mathrm{4}\:\Rightarrow\:{a}=\pm\mathrm{1},\:\pm\mathrm{2},\:\pm\mathrm{3},\:\pm\mathrm{4} \\ $$$$\mid\mathrm{2}{p}+{k}+\mathrm{1}\mid\leqslant\mathrm{4}\wedge\mid{p}^{\mathrm{2}} +{pk}−\mathrm{1}\mid\leqslant\mathrm{4}\:\Rightarrow \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[\mathrm{remember}\:{p}\neq\mathrm{0}\wedge{p}+{k}\neq\mathrm{0}\right] \\ $$$$\:\:\:\:\:{k}=\mathrm{1}\:\Rightarrow\:{p}=−\mathrm{2},\:\mathrm{1} \\ $$$$\:\:\:\:\:{k}=\mathrm{2}\:\Rightarrow\:{p}=−\mathrm{3},\:−\mathrm{1} \\ $$$$\:\:\:\:\:{k}=\mathrm{3}\:\Rightarrow\:{p}=−\mathrm{4},\:−\mathrm{2},\:−\mathrm{1} \\ $$$$\:\:\:\:\:{k}=\mathrm{4}\:\Rightarrow\:{p}=−\mathrm{3},\:−\mathrm{1} \\ $$$$ \\ $$$${L}_{\mathrm{1}} =\left\{{a},\:{p},\:{p}+\mathrm{1},\:−{a}\left(\mathrm{2}{p}+\mathrm{1}\right),\:{ap}\left({p}+\mathrm{1}\right)\right\} \\ $$$$\:\:\:\:\:{p}=−\mathrm{2} \\ $$$$\:\:\:\:\:{L}_{\mathrm{1}} =\left\{{a},\:−\mathrm{2},\:−\mathrm{1},\:\mathrm{3}{a},\:\mathrm{2}{a}\right\}\:\mathrm{impossible} \\ $$$$\:\:\:\:\:{p}=\mathrm{1} \\ $$$$\:\:\:\:\:{L}_{\mathrm{1}} =\left\{{a},\:\mathrm{1},\:\mathrm{2},\:−\mathrm{3}{a},\:\mathrm{2}{a}\right\}\:\mathrm{impossible} \\ $$$${L}_{\mathrm{2}} =\left\{{a},\:{p},\:{p}+\mathrm{2},\:−\mathrm{2}{a}\left({p}+\mathrm{1}\right),\:{ap}\left({p}+\mathrm{2}\right)\right\} \\ $$$$\:\:\:\:\:{p}=−\mathrm{3} \\ $$$$\:\:\:\:\:{L}_{\mathrm{2}} =\left\{{a},\:−\mathrm{3},\:−\mathrm{1},\:\mathrm{4}{a},\:\mathrm{3}{a}\right\}\:\mathrm{impossible} \\ $$$$\:\:\:\:\:{p}=−\mathrm{1} \\ $$$$\:\:\:\:\:{L}_{\mathrm{2}} =\left\{{a},\:−\mathrm{1},\:\mathrm{1},\:\mathrm{0},\:−{a}\right\}\:{a}=\pm\mathrm{2}\:\bigstar \\ $$$${L}_{\mathrm{3}} =\left\{{a},\:{p},\:{p}+\mathrm{3},\:−{a}\left(\mathrm{2}{p}+\mathrm{3}\right),\:{ap}\left({p}+\mathrm{3}\right)\right\} \\ $$$$\:\:\:\:\:{p}=−\mathrm{4} \\ $$$$\:\:\:\:\:{L}_{\mathrm{3}} =\left\{{a},\:−\mathrm{4},\:−\mathrm{1},\:\mathrm{5}{a},\:\mathrm{4}{a}\right\}\:\mathrm{impossible} \\ $$$$\:\:\:\:\:{p}=−\mathrm{2} \\ $$$$\:\:\:\:\:{L}_{\mathrm{3}} =\left\{{a},\:−\mathrm{2},\:\mathrm{1},\:{a},\:−\mathrm{2}{a}\right\}\:\mathrm{impossible} \\ $$$$\:\:\:\:\:{p}=−\mathrm{1} \\ $$$$\:\:\:\:\:{L}_{\mathrm{3}} =\left\{{a},\:−\mathrm{1},\:\mathrm{2},\:−{a},\:−\mathrm{2}{a}\right\}\:\mathrm{impossible} \\ $$$${L}_{\mathrm{4}} =\left\{{a},\:{p},\:{p}+\mathrm{4},\:−\mathrm{2}{a}\left({p}+\mathrm{2}\right),\:{ap}\left({p}+\mathrm{4}\right)\right\} \\ $$$$\:\:\:\:\:{p}=−\mathrm{3} \\ $$$$\:\:\:\:\:{L}_{\mathrm{4}} =\left\{{a},\:−\mathrm{3},\:\mathrm{1},\:\mathrm{2}{a},\:−\mathrm{3}{a}\right\}\:\mathrm{impossible} \\ $$$$\:\:\:\:\:{p}=−\mathrm{1} \\ $$$$\:\:\:\:\:{L}_{\mathrm{4}} =\left\{{a},\:−\mathrm{1},\:\mathrm{3},\:−\mathrm{2}{a},\:−\mathrm{3}{a}\right\}\:\mathrm{impossible} \\ $$$$ \\ $$$$\bigstar \\ $$$$\mathrm{We}\:\mathrm{get}\:\mathrm{2}\:\mathrm{solutions} \\ $$$$\mathrm{1}.\:{a}=−\mathrm{2}\wedge{b}=\mathrm{0}\wedge{c}=\mathrm{2}\wedge{p}=−\mathrm{1}\wedge{q}=\mathrm{1} \\ $$$$−\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{2}.\:{a}=\mathrm{2}\wedge{b}=\mathrm{0}\wedge{c}=−\mathrm{2}\wedge{p}=−\mathrm{1}\wedge{q}=\mathrm{1} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}=\mathrm{0} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *