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# Help

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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
+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
+111396
+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

Jun 26, 2019
edited by CPhill  Jun 26, 2019
#3
+886
+9

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