What is the value of c so that y = x² + 9x + c is a perfect square trinomial?
didn't I just do this?
Sorry, I missed it. Thanks.
To find the value of c such that the quadratic trinomial y = x² + 9x + c is a perfect square, we can use the concept of completing the square.
A perfect square trinomial can be written as the square of a binomial. In this case, the square of the binomial will have the form (x + p)², where p represents a constant.
Let's work through the steps:
Step 1: Rewrite the given quadratic in the form (x + p)² + q, where p and q are constants.
y = x² + 9x + c
Step 2: To find the constant q, take half of the coefficient of x, square it, and add it to both sides of the equation.
y + (9/2)² = x² + 9x + (9/2)² + c
Simplifying:
y + 81/4 = (x + 9/2)² + c
Step 3: Now we have a perfect square on the right side, so the trinomial is a perfect square when the constant term on the left side is zero. Hence, q = -81/4.
-81/4 = c
Therefore, the value of c that makes y = x² + 9x + c a perfect square trinomial is -81/4.