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What constant will make y^2 + 28y + 16 + k the square of a binomial?

 May 19, 2022
 #1
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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?

 May 19, 2022
 #2
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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  

.

 May 20, 2022

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