What is the value of n so that the expression x^2+11x+n Is a perfect square trinomial
To find the value of n, we need to determine what value of n will make the expression a perfect square trinomial.
For a trinomial to be a perfect square, the coefficient of the linear term must be half of the coefficient of the squared term, and the constant term must be the square of half of the coefficient of the linear term.
In this case, the linear term is 11x and the squared term is x^2. So, the coefficient of the linear term is 11 and the coefficient of the squared term is 1.
The coefficient of the linear term (11) is twice the coefficient of the squared term (1), so it satisfies the first condition.
To find the value of n, we square half of the coefficient of the linear term (11/2) to get the constant term.
(11/2)^2 = 121/4
Therefore, the value of n that makes the expression x^2 + 11x + n a perfect square trinomial is 121/4.