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

Feb 15, 2018

$4 {x}^{2} + 32 x + 55$, when expressed in the form
${\left(a x + b\right)}^{2} + c$, we have
$4 {x}^{2} + 32 x + 55 = {\left(2 x + 8\right)}^{2} - 9$

#### Explanation:

$4 {x}^{2} + 32 x + 55$
Here,
Changing
$4 {x}^{2} \to {\left(2 x\right)}^{2}$
$32 x \to 2 \left(2 x\right) \left(8\right)$
we have
$4 {x}^{2} + 32 x + 55 = {\left(2 x\right)}^{2} + 2 \left(2 x\right) \left(8\right) + 55$
We have,
${\left(2 x\right)}^{2} + 2 \left(2 x\right) \left(8\right) + {8}^{2} = {\left(2 x + 8\right)}^{2}$

Since ${8}^{2} = 64$

${\left(2 x\right)}^{2} + 2 \left(2 x\right) \left(8\right) + 64 = {\left(2 x + 8\right)}^{2}$
and
changing now,
$55 = 64 - 9$+We have
$4 {x}^{2} + 32 x + 55 = {\left(2 x\right)}^{2} + 2 \left(2 x\right) \left(8\right) + 64 - 9$
$4 {x}^{2} + 32 x + 55 = {\left(2 x + 8\right)}^{2} - 9$

Feb 15, 2018

Is the question correct? The normal format for this question type is completing the square $\to a {\left(x + b\right)}^{2} + c$
Answering for $a {\left(x + b\right)}^{2} + c$

$y = 4 {\left(x + 4\right)}^{2} - 9$

#### Explanation:

$\textcolor{b l u e}{\text{Preamble}}$

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

$2 = 2 \leftarrow \text{ Original structure}$

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

$3 - 1 = 2 \leftarrow \text{ Now we have 'forced' it to be true}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Answering the question}}$

Given $\underbrace{y = 4 {x}^{2} + 32 x + 55 \leftarrow} \text{ Starting point}$
$\textcolor{w h i t e}{\text{ddddddddddddd}} \downarrow$

$\underbrace{\text{You should go directly from that to this:}}$
$\textcolor{w h i t e}{\text{dddddddddddddd}} \downarrow$

$\textcolor{w h i t e}{\text{ddddd}} y = 4 {\left(x + 4\right)}^{2} + k + 55$

then determine the value of $k$

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

$\textcolor{b r o w n}{\text{This is how I got there taking it one step at a time.}}$

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.

$\textcolor{b l u e}{S t e p \textcolor{w h i t e}{\text{d}} 1 \implies {k}_{1}}$

$y = \left(4 {x}^{2} + 32 x\right) + {k}_{1} + 55$ where in the end $k + 55 \to c$
At this point ${k}_{1} = 0$ as we have not changed any values.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{S t e p 2 \implies {k}_{1}}$ No values changed yet

Factor out the 4

$y = 4 \left({x}^{2} + \frac{32}{4} x\right) + {k}_{1} + 55$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b l u e}{S t e p \textcolor{w h i t e}{\text{d}} 3 \implies {k}_{2}}$ Values starting to change: Halve the $\frac{32}{4} x$

$y = 4 \left({x}^{2} + \frac{32}{8} x\right) + {k}_{2} + 55$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b l u e}{S t e p \textcolor{w h i t e}{\text{d}} 4 \implies {k}_{3}}$ Remove the $x$ from $\frac{32}{8} x$

$y = 4 \left({x}^{2} + \frac{32}{8}\right) + {k}_{3} + 55$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\textcolor{b l u e}{S t e p \textcolor{w h i t e}{\text{d}} 5 \implies {k}_{4}}$ Move the 2 from ${x}^{2}$ to outside the brackets

$y = 4 {\left(x + \frac{32}{8}\right)}^{2} + {k}_{4} + 55$

$y = 4 {\left(x + 4\right)}^{2} + {k}_{4} + 55 \leftarrow \text{ Nearly there!}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{S t e p \textcolor{w h i t e}{\text{d}} 6 \implies {k}_{4}}$ Determine the value of ${k}_{4}$

$y = \textcolor{g r e e n}{4} {\left(x + \textcolor{m a \ge n t a}{4}\right)}^{2} + {k}_{4} + 55$

Set $\left(\textcolor{g r e e n}{4} \times \textcolor{m a \ge n t a}{{4}^{2}}\right) + {k}_{4} = 0 \leftarrow \text{ The correction}$

$64 + {k}_{4} = 0 \implies {k}_{4} = - 64$ So substituting for ${k}_{4}$ we have:

$y = 4 {\left(x + 4\right)}^{2} + {k}_{4} + 55 \textcolor{w h i t e}{\text{ddd")->color(white)("dddd}} y = 4 {\left(x + 4\right)}^{2} - 64 + 55$

$\textcolor{w h i t e}{\text{dddddddddddddddddddddd")-> color(white)("dddd}} y = 4 {\left(x + 4\right)}^{2} - 9$

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

Feb 15, 2018

'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:

$y = {\left(2 x + 8\right)}^{2} - 9$

#### Explanation:

Expanding : $y = {\left(2 x + 8\right)}^{2} - 9$

$y = 4 {x}^{2} + 32 x + 64 - 9$

$y = 4 {x}^{2} + 32 x + 55$