For what real values of c is x^2 + 16x + c the square of a binomial? If you find more than one, then list your values in increasing order, separated by commas.
completing the square
(16 / 2)² = 64
(x + 8)²
Did you mean
(16/2)^2=64(x+8)^2?
To determine the real values of c for which the given quadratic expression, x^2 + 16x + c, is the square of a binomial, we need to find the perfect square trinomial it corresponds to and derive its factors.
A perfect square trinomial can be written in the form (x + a)^2, where a is a constant value. Expanding this expression will give us x^2 + 2ax + a^2. By comparing this with the given quadratic expression x^2 + 16x + c, we can identify that the constant term c must be a perfect square to match the pattern.
Therefore, we need to find the perfect squares that are greater than or equal to zero (as c should be a real value). We can start by listing the perfect squares in increasing order: 0, 1, 4, 9, 16, 25, ...
From the list, we can observe that the perfect square values for c that satisfy the given criteria are 0 and 16. These values will result in the quadratic expression being the square of a binomial. Therefore, the real values of c for which x^2 + 16x + c is the square of a binomial are 0 and 16.