IF (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......
Answers
Answered by
47
hello friend
This is a very interesting question.
A^2 + B^2 = 0
It means A=B=0
Because after square any positive and negative number ..Will become positive number and the sum of positive numbers can not be equal to zero ...
So the sum will be zero only for the square numbers individually zero
a-5 = 0
a = 5...(1)
b-c=0
b=c....(2)
C=d...(3)
b+c+d=9...(4)
Form equation 2 and 3 and 4
b=c=d=3
Put all the value in given equation
(a+b+c)(b+c+d) =
(5+3+3)(9) = 99
Hope it will help you !!
This is a very interesting question.
A^2 + B^2 = 0
It means A=B=0
Because after square any positive and negative number ..Will become positive number and the sum of positive numbers can not be equal to zero ...
So the sum will be zero only for the square numbers individually zero
a-5 = 0
a = 5...(1)
b-c=0
b=c....(2)
C=d...(3)
b+c+d=9...(4)
Form equation 2 and 3 and 4
b=c=d=3
Put all the value in given equation
(a+b+c)(b+c+d) =
(5+3+3)(9) = 99
Hope it will help you !!
karunarandive:
par yaha A aur B ky hai
Yeh concept explain karne k liye use kiya hai ... A and B koi bhi number assume kar sakte ho .. positive negetive ... Inke square karke ..Sum karoge to kabhi zero answer nahi ayega ....
us question me a tumne kya assume kiya hai
aur A Aur B kya assume kiya hai
A and B koi bhi number ho sakte hai ..Like 2,(-3) , 4 ,(-10) ,
Above question me tumne is formula ka application kiya hai na to mere question me A aur B kya assume kiya hai
Ok... Confusion hai yaha shayad
A^2 +B^2 +C^2 +.......+z^2 =0 , kitne bhi terms ho question me ... Yaha question me 4 terms hai ... A=B=C=D=...=Z=0
A= a-5 ,. B=b-c ..., C= c-d ... ,D= b+c+d .....
May be ..Ab clear ho gaya hoga point ??.. if u have any doubt please ask .. I would like to help you.
Answered by
15
Answer :-
→ 99 .
Step-by-step explanation :-
°•° (a – 5)² + (b – c)² + (c – d)² + (b + c + d – 9)² = 0 .
Therefore,
==> a – 5 = 0 or a = 5 ... (i) .
==> b – c = 0 or b = c ...(ii) .
==> c – d = 0 or c = d...(iii) .
==> b + c + d – 9 = 0 ... (iv) .
Now,
•°• b = c = d ...(v). [ From equation (ii) and (iii) . ]
So, by equation (ii) and (iii), we get
==> 3b = 9.
•°• b = 3.
So,
•°• a = 5, b = 3, c = 3 and d = 3 .
Now,
= ( a + b + c )( b + c + d ) .
= (5 + 3 + 3) (3 + 3 + 3).
= (11) (9) .
= 99 .
Hence, it is solved.
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