Question

if (a-5)^2 +(b-c)^2 +(c-d)^2+(b+c+d-9)^2 =0 then what will be (a+b+c)(b+c+d) =?

Answer 1

here (a-5)^2 +(b-c)^2 +(c-d)^2+(b+c+d-9)^2 =0
all the terms are square so no term can be negative. Their sum is zero means that each term is zero in itself.
(a-5) to the power of 2 = 0 it implies a=5
(b-c) to the power of 2 = 0 it implies b=c
(c-d) to the power of 2 = 0 it implies c=d
this means that b=c=d. Now replacing the values in the equation, we get
(b+c+d-9)whole square = 0 it implies (3b-9)whole square = 0 it implies b=3
this means that b=c=d=3
now putting the value in a,b,c and d in eq.(a+b+c)(b+c+d) we get
= (5+3+3)(3+3+3)
= 11 x 9 = 99

hopw this would have cleared your doubt.

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