Is:
n^2-9n+9 a perfect square??
No
Can you please explain how you got this, so I can understand it better?
(n-3)(n-3) = n^2 -6 n + 9
unfortunately that 6 is not a 9
Now we could try completing the square
n^2 - 9 n + 9 = y
n^2 - 9 n = y -9
n^2 - 9 n + (9/2)^2 = y - 9 + 81/4
so
n^2 - 9 n + 81/4
is a perfect square :)
(n-9/2)(n-9/2)
To determine if the expression n^2 - 9n + 9 is a perfect square, we need to factor the expression and check if the factors can be written in the form (a - b)^2.
Let's start by factoring the expression:
n^2 - 9n + 9
We can rewrite -9n as -3n - 6n:
n^2 - 3n - 6n + 9
Now, we can group the terms and factor them separately:
(n^2 - 3n) - (6n - 9)
n(n - 3) - 3(2n - 3)
Now, we have two binomials: (n - 3) and (2n - 3).
If the given expression is a perfect square, both of these binomials should have the same factors. Let's check if that's the case:
The first binomial (n - 3) is already factored, so we can leave it as it is.
For the second binomial (2n - 3), we can check if it can be written in the form (a - b)^2 by applying the perfect square formula: (a - b)^2 = a^2 - 2ab + b^2.
Comparing this with our binomial (2n - 3), we can see that "a" would be 2n and "b" would be √3. However, the coefficient of "ab" is -2ab in the perfect square formula, while in our binomial it is -6, which is not equal to -2. Therefore, (2n - 3) cannot be factored as a perfect square.
Since the binomials do not have the same factors, we can conclude that n^2 - 9n + 9 is not a perfect square.