Question

Write an equation equivalent to the one below by writing  the trinomial as a perfect square trinomial. `x^2 ­ - 4x + 1 = 0` ?

A: `x^2 - 4x + 4 = -3`

B: `x^2 - ­ 4x + 5 = 0`

C: `x^2 ­- 4x + 4 = ­3`

D: `x^2 ­- 4x + 4 = ­5`

Answer 1

C

Explanation:

Look at https://socratic.org/s/aNNKeJ73 for an in-depth explanation of the steps for completing the square,

Given `x^2-4x+1=0`

half of the 4 from `-4x` is 2 so we have

`(xcolor(red)(-2))^2+k+1=0` where `k` is some constant

Set `(color(red)(-2))^2+k=0 => k=-4`

Thus we have

`(x-2)^2-4+1=0`

`ubrace(color(white)("d")(x-2)^2color(white)("d"))color(white)("ddd")-3=0 larr" Completing the square"`
`x^2-4x+4color(white)("dd")-3=0`

Add `3` to both sides

`x^2+4x+4=3 larr" Option C"color(red)( larr "Corrected from option D")`

Answer 2

Option `C`

Explanation:

This is by a process known as 'completing the square'

You need to add in a missing value so that you have a trinomial which is a perfect square.

The missing term is `(b/2)^2`

`x^2 -color(blue)(4)x +1 =0" "larr (b=color(blue)(-4))`

`x^2 -4x + ((color(blue)(-4))/2)^2 = -1+((color(blue)(-4))/2)^2`

`x^2 -4x +4 = -1+4`

`x^2 -4x +4 = 3`

the left side is now equal to `(x-2)^2`, a perfect square

So option `C` is the one you want.