For the quadratic to be the square of a binomial, the discriminant(\(b^2 -4ac\)) must equal 0.
Subsituting what we know, we have: \(28^2 - 4(16 + k) = 0 \).
Now, we just have to solve for k.
Can you take it from here?
What constant will make y^2 + 28y + 16 + k the square of a binomial?
This one is easy because the coefficient of the squared term is 1.
Remember that in general terms (x + a)2 = x2 + 2ax + a2
So, for y2 + 28y + a2 the value represented by 2a is 28
You just take half of 28 to get "a" a = 28/2 = 14
Of course, you want a2 so square 14 to get 196
But you already had 16 so k = 196 – 16 thus k = 180
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