Find the value of n such that x^2 - 10x + n is a perfect square trinomial.
To find the value of n such that x^2 - 10x + n is a perfect square trinomial, we need to determine the value of n such that the trinomial can be factored into the form (x - a)^2, where a is a constant.
Expanding (x - a)^2, we get x^2 - 2ax + a^2.
Comparing this with x^2 - 10x + n, we can set up the following equations:
-2a = -10 (comparing the coefficients of x)
a^2 = n (comparing the constant terms)
From the first equation, we can solve for a:
-2a = -10
a = 5
Substituting this value of a into the second equation, we can solve for n:
a^2 = n
5^2 = n
25 = n
Therefore, the value of n such that the trinomial x^2 - 10x + n is a perfect square trinomial is 25.