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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 =?
onesolutionoftheequation(xa)(xb)(xc)(xd)=9isx=2.Ifa,b,c,daredifferentintegersthena+b+c+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✓
(xa)(xb)(xc)(xd)=9x=2x=2(2a)(2b)(2c)(2d)=9a,b,c,daredifferentintegers2a,2b,2c,2daredifferentintegers.9=1×1×3×3[Ordernomatters]2a=1a=12b=1b=32c=3c=12d=3d=5a+b+c+d=8
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  ×
letm=a+b+c+d&2a=p&2b=q&2c=r&2d=sm=8(p+q+r+s)1)p=1,q=1,r=3,s=3m=0×2)p=1,q=1,r=3,s=3m=4×3)p=1,q=1,r=3,s=3m=12×4)p=1,q=1,r=3,s=3m=85)p=1,q=1,r=3,s=3m=15×6)p=1,q=1,r=1,s=9m=4×7)p=1,q=1,r=1,s=9m=0×8)p=1,q=1,r=1,s=9m=16×9)p=1,q=1,r=1,s=9m=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 ✓
saya<b<c<d(a2)(b2)(c2)(d2)=9letp=a2,q=b2,r=c2,s=d2a+b+c+d=p+q+r+s+8pqrs=9withp<q<r<sonlyonepossibility:pqrs=(3)(1)(1)(3)=9a+b+c+d=31+1+3+8=8
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.
youareright.ididnopayattentiontothedifferenceinnumberThankyousir.

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