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Question Number 196450 by cortano12 last updated on 25/Aug/23

one solution of the equation (x−a)(x−b)(x−c)(x−d) = 9 is x=2. If a,b,c,d are different integers then a+b+c+d =?

$$\mathrm{one}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\: \\ $$$$\:\left(\mathrm{x}−\mathrm{a}\right)\left(\mathrm{x}−\mathrm{b}\right)\left(\mathrm{x}−\mathrm{c}\right)\left(\mathrm{x}−\mathrm{d}\right)\:=\:\mathrm{9}\: \\ $$$$\:\mathrm{is}\:\mathrm{x}=\mathrm{2}.\:\mathrm{If}\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\:\mathrm{are}\:\mathrm{different}\: \\ $$$$\:\mathrm{integers}\:\mathrm{then}\:\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}\:=?\: \\ $$

Answered by Rasheed.Sindhi last updated on 25/Aug/23

(x−a)(x−b)(x−c)(x−d) = 9 x=2 x=2 (2−a)(2−b)(2−c)(2−d)=9 a,b,c,d are different integers ⇒2−a,2−b,2−c,2−d are different integers. • 9=1×−1×3×−3 [Order no matters] 2−a=1⇒a=1 2−b=−1⇒b=3 2−c=3⇒c=−1 2−d=−3⇒d=5 a+b+c+d=8✓

$$\left(\mathrm{x}−\mathrm{a}\right)\left(\mathrm{x}−\mathrm{b}\right)\left(\mathrm{x}−\mathrm{c}\right)\left(\mathrm{x}−\mathrm{d}\right)\:=\:\mathrm{9} \\ $$$$\mathrm{x}=\mathrm{2}\: \\ $$$$\mathrm{x}=\mathrm{2} \\ $$$$\left(\mathrm{2}−\mathrm{a}\right)\left(\mathrm{2}−\mathrm{b}\right)\left(\mathrm{2}−\mathrm{c}\right)\left(\mathrm{2}−\mathrm{d}\right)=\mathrm{9} \\ $$$$\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\:\mathrm{are}\:\mathrm{different}\:\mathrm{integers} \\ $$$$\Rightarrow\mathrm{2}−\mathrm{a},\mathrm{2}−\mathrm{b},\mathrm{2}−\mathrm{c},\mathrm{2}−\mathrm{d}\:\mathrm{are}\:\mathrm{different}\:\mathrm{integers}. \\ $$$$\: \\ $$$$\bullet\:\mathrm{9}=\mathrm{1}×−\mathrm{1}×\mathrm{3}×−\mathrm{3}\:\left[{Order}\:{no}\:{matters}\right] \\ $$$$\mathrm{2}−\mathrm{a}=\mathrm{1}\Rightarrow\mathrm{a}=\mathrm{1} \\ $$$$\mathrm{2}−\mathrm{b}=−\mathrm{1}\Rightarrow\mathrm{b}=\mathrm{3} \\ $$$$\mathrm{2}−\mathrm{c}=\mathrm{3}\Rightarrow\mathrm{c}=−\mathrm{1} \\ $$$$\mathrm{2}−\mathrm{d}=−\mathrm{3}\Rightarrow\mathrm{d}=\mathrm{5} \\ $$$$\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=\mathrm{8}\checkmark \\ $$

Answered by MM42 last updated on 25/Aug/23

let m=a+b+c+d & 2−a=p & 2−b=q & 2−c=r & 2−d=s ⇒m=8−(p+q+r+s) 1) p=1 , q=1 ,r=3 , s=3 ⇒m=0 × 2)p=−1 , q=−1 ,r=3 ,s=3 ⇒m=4 × 3)p=1 , q=1 ,r=−3 ,s=−3 ⇒m=12 × 4)p=1 , q=−1 ,r=3 ,s=−3 ⇒m=8 ✓ 5)p=−1 , q=−1 ,r=−3 ,s=−3 ⇒m=15 × 6)p=1 , q=1 ,r=1 ,s=9 ⇒m=−4 × 7)p=−1 , q=−1 ,r=1 ,s=9 ⇒m=0 × 8)p=1 , q=1 ,r=−1 ,s=−9 ⇒m=16 × 9)p=−1 , q=−1 ,r=−1 ,s=−9 ⇒m=20 ×

$${let}\:{m}={a}+{b}+{c}+{d}\:\:\:\:\&\:\:\mathrm{2}−{a}={p}\:\:\&\:\:\mathrm{2}−{b}={q}\:\:\&\:\:\mathrm{2}−{c}={r}\:\:\&\:\:\mathrm{2}−{d}={s} \\ $$$$\Rightarrow{m}=\mathrm{8}−\left({p}+{q}+{r}+{s}\right) \\ $$$$\left.\mathrm{1}\right)\:{p}=\mathrm{1}\:\:,\:{q}=\mathrm{1}\:\:,{r}=\mathrm{3}\:\:,\:{s}=\mathrm{3}\:\Rightarrow{m}=\mathrm{0}\:× \\ $$$$\left.\mathrm{2}\right){p}=−\mathrm{1}\:\:,\:{q}=−\mathrm{1}\:\:,{r}=\mathrm{3}\:\:,{s}=\mathrm{3}\:\Rightarrow{m}=\mathrm{4}\:\:× \\ $$$$\left.\mathrm{3}\right){p}=\mathrm{1}\:\:,\:{q}=\mathrm{1}\:\:,{r}=−\mathrm{3}\:\:,{s}=−\mathrm{3}\:\Rightarrow{m}=\mathrm{12}\:\:× \\ $$$$\left.\mathrm{4}\right){p}=\mathrm{1}\:\:,\:{q}=−\mathrm{1}\:\:,{r}=\mathrm{3}\:\:,{s}=−\mathrm{3}\:\Rightarrow{m}=\mathrm{8}\:\:\checkmark \\ $$$$\left.\mathrm{5}\right){p}=−\mathrm{1}\:\:,\:{q}=−\mathrm{1}\:\:,{r}=−\mathrm{3}\:\:,{s}=−\mathrm{3}\:\Rightarrow{m}=\mathrm{15}\:\:× \\ $$$$\left.\mathrm{6}\right){p}=\mathrm{1}\:\:,\:{q}=\mathrm{1}\:\:,{r}=\mathrm{1}\:\:,{s}=\mathrm{9}\:\Rightarrow{m}=−\mathrm{4}\:\:× \\ $$$$\left.\mathrm{7}\right){p}=−\mathrm{1}\:\:,\:{q}=−\mathrm{1}\:\:,{r}=\mathrm{1}\:\:,{s}=\mathrm{9}\:\Rightarrow{m}=\mathrm{0}\:\:× \\ $$$$\left.\mathrm{8}\right){p}=\mathrm{1}\:\:,\:{q}=\mathrm{1}\:\:,{r}=−\mathrm{1}\:\:,{s}=−\mathrm{9}\:\Rightarrow{m}=\mathrm{16}\:\:× \\ $$$$\left.\mathrm{9}\right){p}=−\mathrm{1}\:\:,\:{q}=−\mathrm{1}\:\:,{r}=−\mathrm{1}\:\:,{s}=−\mathrm{9}\:\Rightarrow{m}=\mathrm{20}\:\:× \\ $$$$ \\ $$

Answered by mr W last updated on 25/Aug/23

say a<b<c<d (a−2)(b−2)(c−2)(d−2)=9 let p=a−2, q=b−2, r=c−2, s=d−2 a+b+c+d=p+q+r+s+8 pqrs=9 with p<q<r<s only one possibility: pqrs=(−3)(−1)(1)(3)=9 ⇒a+b+c+d=−3−1+1+3+8=8 ✓

$${say}\:{a}<{b}<{c}<{d} \\ $$$$\left({a}−\mathrm{2}\right)\left({b}−\mathrm{2}\right)\left({c}−\mathrm{2}\right)\left({d}−\mathrm{2}\right)=\mathrm{9} \\ $$$${let}\:{p}={a}−\mathrm{2},\:{q}={b}−\mathrm{2},\:{r}={c}−\mathrm{2},\:{s}={d}−\mathrm{2} \\ $$$${a}+{b}+{c}+{d}={p}+{q}+{r}+{s}+\mathrm{8} \\ $$$${pqrs}=\mathrm{9} \\ $$$${with}\:{p}<{q}<{r}<{s} \\ $$$${only}\:{one}\:{possibility}: \\ $$$${pqrs}=\left(−\mathrm{3}\right)\left(−\mathrm{1}\right)\left(\mathrm{1}\right)\left(\mathrm{3}\right)=\mathrm{9} \\ $$$$\Rightarrow{a}+{b}+{c}+{d}=−\mathrm{3}−\mathrm{1}+\mathrm{1}+\mathrm{3}+\mathrm{8}=\mathrm{8}\:\checkmark \\ $$

Commented by MM42 last updated on 25/Aug/23

you are right.i did no pay attention to the difference in number Thank you sir.

$${you}\:{are}\:{right}.{i}\:{did}\:{no}\:{pay}\:{attention}\: \\ $$$${to}\:{the}\:{difference}\:{in}\:{number} \\ $$$${Thank}\:{you}\:{sir}. \\ $$$$ \\ $$

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