For what real values of c is 9x^2 + 16x + c the square of a binomial? If you find more than one, then list your values in increasing order, separated by commas.
do you know about completing the square?
(3x + y)(3x+y)= 9 x^2 + 6xy + y^2
6 y = 16
y = 8/3
y^2 = 64/9
so
9 x^2 + 16 x + 64/9
To determine the values of "c" for which the polynomial 9x^2 + 16x + c is the square of a binomial, we can use the concept of perfect square trinomials.
A perfect square trinomial is in the form (a + b)^2, where "a" and "b" are constants. When we expand (a + b)^2, we get a^2 + 2ab + b^2, which is a trinomial.
In order for 9x^2 + 16x + c to be a perfect square trinomial, it must be in the form (3x + d)^2, where "d" is a constant. Expanding (3x + d)^2, we get 9x^2 + 6dx + d^2.
Comparing this with the given expression 9x^2 + 16x + c, we see that 6dx = 16x, and d^2 = c.
First, let's solve 6dx = 16x for "d":
6dx = 16x
6d = 16 (dividing both sides by x)
d = 16/6
d = 8/3
Now, let's substitute d = 8/3 into d^2 = c:
(8/3)^2 = c
64/9 = c
So, the value of "c" is 64/9.
Therefore, the only real value of "c" for which 9x^2 + 16x + c is the square of a binomial is c = 64/9.