Gate
Exam
(a-5)2+(b-c)2+(c-d)2+(b+c+d-9)2=0
then the value of (a+b+c)(b+c+d) is ?
a)0
b)11
c)33
d)99
Read Solution (Total 1)
-
- (a-5)2+(b-c)2+(c-d)2+(b+c+d-9)2=0
=> (a-5)2=0,
(b-c)2=0, (c-d)2=0, (b+c+d-9)2=0
=> a=5,b=c,c=d,b+c-d=9
=> b=c=d and b+c+d=9
hence b+b+b=9
=> b=3
so c=d=3
and we have a=5
putting all the values in (a+b+c)(b+c+d)
we get (5+3+3)(3+3+3)=11*9=99 - 11 years agoHelpfull: Yes(1) No(1)
Gate Other Question