#1**0 **

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?

BuilderBoi May 19, 2022

#2**0 **

*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} = x^{2} + 2ax + a^{2}

So, for y^{2} + 28y + a^{2} the value represented by 2a is 28

You just take half of 28 to get "a" a = 28/2 = **14**

Of course, you want a^{2} so square 14 to get 196

But you already had 16 so k = 196 – 16 thus **k = 180**

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Guest May 20, 2022