Write x2 − 6x + 7 = 0 in the form (x − a)2 = b, where a and b are integers.
a). (x − 4)2 = 3
b). (x − 3)2 = 2
c). (x − 2)2 = 1
d). (x − 1)2 = 4

X² − 6x + 7 = 0

Let's complete the square [x²-6x+(??)²] -(??)² +7 = 0

As you noticed we have added the unknown square & also subtract the same so that not to change the equation

The question now is to calculate this square (??)²

We know that the middle term (6x) is twice the square root of the 1st times the square root of the 2nd, That is 6x=2x (multiply) √(??)²

6x = 2x?? & ??=6x/2x===> ?? =3 & (??)² =9

Hence plug (??)² into the equation above:

x² - 6x + 9 - 9 +7 =0

(x²-6x+9) -9 + 7 =0

(x-3)² -2 =0 > (x-3)² = 2 (that is B)

A. The order of objects counted in a permutation always matters.

Step-by-step explanation:

Option A is the right answer.

The order of objects counted in a permutation always matters.

Permutation are used for lists where order always matter. Permutation is used to arrange members of a set into an order. With permutation, every little detail matters. The number of permutations on a set of 'n' elements is given by n!