Question

Express `4x^2+32x+55` in the form `(ax+b)^2+c`, where a, b and c are constants and a is positive?

Answer 1

`4x^2+32x+55`, when expressed in the form
`(ax+b)^2+c`, we have
`4x^2+32x+55=(2x+8)^2-9`

Explanation:

`4x^2+32x+55`
Here,
Changing
`4x^2 to (2x)^2`
`32x to2(2x)(8)`
we have
`4x^2+32x+55=(2x)^2+2(2x)(8)+55`
We have,
`(2x)^2+2(2x)(8)+8^2=(2x+8)^2`

Since `8^2=64`

`(2x)^2+2(2x)(8)+64=(2x+8)^2`
and
changing now,
`55=64-9`+We have
`4x^2+32x+55=(2x)^2+2(2x)(8)+64-9`
`4x^2+32x+55=(2x+8)^2-9`

Answer 2

Is the question correct? The normal format for this question type is completing the square `->a(x+b)^2+c`
Answering for `a(x+b)^2+c`

`y=4(x+4)^2-9`

Explanation:

`color(blue)("Preamble")`

You can change any equation you wish into the format you wish as long as you add to it a correction that if applied with transform it back again.

A bit like

`2=2larr" Original structure"`

`3=2 larr" Changed and "color(red)("not true")`

`3-1=2 larr" Now we have 'forced' it to be true"`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Answering the question")`

Given `ubrace(y=4x^2+32x+55 larr)" Starting point""`
`color(white)("ddddddddddddd")darr`

`ubrace("You should go directly from that to this:")`
`color(white)("dddddddddddddd")darr`

`color(white)("ddddd")y=4(x+4)^2+k+55`

then determine the value of `k`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(brown)("This is how I got there taking it one step at a time.")`

I was challenged once on a point so I am doing it this way:

For each change we must include a correction. The value of this correction will be different as we proceed step-wise through the modifications on the path to the final format.

`color(blue)(Stepcolor(white)("d") 1 => k_1)`

`y=(4x^2+32x)+k_1+55` where in the end `k+55->c`
At this point `k_1=0` as we have not changed any values.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)(Step 2 => k_1)` No values changed yet

Factor out the 4

`y=4(x^2+32/4x)+k_1+55`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)(Stepcolor(white)("d") 3 => k_2)` Values starting to change: Halve the `32/4x`

`y=4(x^2+32/8x)+k_2+55`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)(Stepcolor(white)("d")4 => k_3)` Remove the `x` from `32/8x`

`y=4(x^2+32/8)+k_3+55`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

`color(blue)(Stepcolor(white)("d")5 => k_4)` Move the 2 from `x^2` to outside the brackets

`y=4(x+32/8)^2+k_4+55`

`y=4(x+4)^2+k_4+55 larr" Nearly there!"`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)(Stepcolor(white)("d")6 => k_4)` Determine the value of `k_4`

`y=color(green)(4)(x+color(magenta)(4))^2+k_4+55`

Set `(color(green)(4)xxcolor(magenta)(4^2) )+k_4=0 larr" The correction"`

`64+k_4=0=>k_4=-64` So substituting for `k_4` we have:

`y=4(x+4)^2+k_4+55 color(white)("ddd")->color(white)("dddd") y=4(x+4)^2-64+55`

`color(white)("dddddddddddddddddddddd")-> color(white)("dddd") y=4(x+4)^2-9`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Tony B

Answer 3

Assuming your question is correct.

'Forcing' the completed square solution in my other answer into this format we have:

`y=(sqrt(4)color(white)("d")x+4sqrt(4)color(white)(.))^2-9" "` which gives:

`y=(2x+8)^2-9`

Explanation:

Expanding : `y=(2x+8)^2-9`

`y=4x^2+32x+64-9`

`y=4x^2+32x+55`

Tony B