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If $a$ is a constant such that $4x^2 - 12x + a$ is the square of a binomial, then what is $a$?

 Jun 26, 2019
 #1
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+1

 

I don't understand the question, but (2x – 3)2 = 4x2 – 12x + 9 if your answer is in there somewhere. 

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 Jun 26, 2019
 #2
avatar+104962 
+2

4x^2 - 12x +  a

 

We can write

 

(2x - m)^2  =   4x^2  - 4mx + m^2

 

This implies that 4m  = 12....so  m = 3     and m^2  = 9   = "a"

 

So we have

 

4x^2 - 12x + 9   =   (2x- 3)^2

 

 

cool cool cool

 Jun 26, 2019
edited by CPhill  Jun 26, 2019
 #3
avatar+778 
+6

So a square of a binomial would be in the form of \((a + b)^2\). In your case, we have \(4x^2 - 12x + a\), which is supposed to be the square of a binomial. 

 

So your binomial would probably be \((2x - 3)\). Squaring this would give us \(4x^2 - 12x + 9\).

 

So, \(\boxed{a = 9}\)

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 Jun 26, 2019

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