# What value of c makes 4x^2+4x+C a perfect square trinomial? a.4 b.1/2 c.1 d.1/4

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`(ax+b)^2 = a^2x^2+2abx+b^2`

`4x^2+4x+C = a^2x^2+2abx+b^2`

Let us compare the components;

`x^2 rarr 4 = a^2 rarr = a = +-2`

`C = b^2`

`x rarr 4 = 2ab `

`4 = 2bxx(+-2)`

When a = +2 then b = 1

When a = -2 then b = -1

`C = b^2` `rarr` `C = 1^2` or `C = (-1)^2`

C = 1

*So the answer is C = 1*

Okay, so to make it a perfect square, you want your equation to look something like this: a² + 2ab + b² = (a+b)² = (a+b) (a+b)

So, lets try to factor the equation:

4x² + 4x + c [you know that 4x² is supposed to be a perfect square, so let's break it down so that it looks like this ( )²]

∴ 4x² = (2x)²

∴ 4x² + 4x + c

= (2x)² + 4x + c [now let's break up the 4x]:

= (2x)² + (2x + 2x) + c

= (2x)² + 2x + 2x + c [now let's factor out what we can]

= 2x (2x + 1) + 1 (2x + c)

okay. so we need to make sure that 2x + c is equal to 2x + 1, so that it can be a perfect square:

2x + c = 2x + 1 [move the 2x's to the left]

2x - 2x + c = 1

∴ c = 1

now check to see if it will work:

2x (2x + 1) + 1 (2x + c) , let c = 1:

= 2x (2x + 1) + 1 (2x + 1)

= (2x + 1) (2x + 1)

= (2x + 1)²

so, it works. Therefore, c = 1.