Question

(a-1)^2+(b-2)^2+(c-3)^2+(d-4)^2=0 then the value of a*b*c*d+1 is

Answer 1

Answer:

W+X+Y+Z=0

Now, if all of W, X, Y and Z are complete squares, which they are according to the question statement, and their sum is 0, we can be sure that all of them are 0s as well, or mathematically put,

W=0, X=0, Y=0, Z=0, if W+X+Y+Z=0 and W, X, Y and Z are perfect squares.

According to the given question,

W=(a−1)2; X=(b−2)2; Y=(c−3)2; Z=(d−4)2

Thus,

W=0⇒(a−1)2=0⇒a−1=0⇒a=1

X=0⇒(b−2)2=0⇒b−2=0⇒b=2

Y=0⇒(c−3)2=0⇒c−3=0⇒c=3

Z=0⇒(d−4)2=0⇒d−4=0⇒d=4

Thus, to answer the question, the value of the expression

abcd+1=1∗2∗3∗4+1=24+1=25

Hope that helped.

</EOA152>

Step-by-step explanation:

please Mark me as brainlest Pleazzzzzz.

Answer 2

Answer:

The final answer is 25.

Step-by-step explanation:

Given,

$(a-1) ^2 +(b-2)^2+(c-3)^2+(d-4)^2=0$.  We need to find the value of the equation a * b * c * d + 1.

Let's substitute these 4 equations into into 4 alphabets.

p = a - 1

q = b - 2

r = c - 3

s = d - 4

Substiting these equations we find out the sum of the square of p, q, r, s is a zero. There is absolutely no way that a square of a number will be a negative number hence p,q,r,s are positive numbers however their sum is zero which is only possible if all of them are equal to 0.

Hence

a - 1 = 0

a = 1

b - 2 = 0

b = 2

c - 3 = 0

c = 3

d - 4 = 0

d = 4

Hence

a * b * c * d + 1 = 1 * 2 * 3 * 4 + 1 = 24 + 1

= 25  

How to substitute values in an equation

https://brainly.in/question/20035006

Similar Problems

https://brainly.in/question/26285864

#SPJ2