Completing the Square: Calculate Derivative of 2x²+14√2x-15 Without Solving for X

First, let's recall the principles of the "completing the square" method and its general idea:

In this method, we use the formulas for the square of a binomial in order to give an expression the form of a squared binomial,

This method is called "completing the square" due to the fact that in this method we "complete" a missing part of a certain expression in order to obtain from it a form of a squared binomial,

That is, we use the formulas for the square of a binomial:

$(c\pm d)^2=c^2\pm2cd+d^2$

And we bring the expression to a squared form by adding and subtracting the missing term,

In the given problem we will first refer to the given equation

$2x^2+14\sqrt{2}x-15=0$

First, we will try to give the expression on the left side of the equation a form that resembles the form of the right side in the abbreviated multiplication formulas mentioned, we will also identify that we are interested in the addition form of the abbreviated multiplication formula, this is because the term that is not squared in the given expression,:

$14\sqrt{2}x$

has a positive sign,we will continue,

First, we will deal with the two terms with the highest powers in the expression requested on the left side of the equation,

And we will try to identify the missing term in comparison to the abbreviated multiplication formula,

To do this- first we will present these terms in a form similar to the form of the first two terms in the abbreviated multiplication formula:

$\underline{ 2x^2+14\sqrt{2}x}-15\textcolor{blue}{\leftrightarrow} \underline{ c^2+2cd+d^2 }\\ \\ \hspace{4pt}\\ \\ \underline{ (\sqrt{2})^2x^2+14\sqrt{2}x}-15\textcolor{blue}{\leftrightarrow} \underline{ c^2+2cd+d^2 }\\ \\ \downarrow\\ \underline{(\textcolor{red}{\sqrt{2}x})^2\stackrel{\downarrow}{+2 }\cdot \textcolor{red}{\sqrt{2}x}\cdot \textcolor{green}{7}}-15 \textcolor{blue}{\leftrightarrow} \underline{ \textcolor{red}{c}^2\stackrel{\downarrow}{+2 }\textcolor{red}{c}\textcolor{green}{d}\hspace{2pt}\boxed{+\textcolor{green}{d}^2}} \\$

It can be noticed that in comparison to the abbreviated multiplication formula (which is on the right side of the blue arrow in the previous calculation) we are actually making the analogy:

$\begin{cases} \sqrt{2}x\textcolor{blue}{\leftrightarrow}c\\ 7\textcolor{blue}{\leftrightarrow}d \end{cases}$

Therefore, we identify that if we want to get a squared binomial form from these two terms (underlined below in the calculation),

we will need to add to these two terms the term $7^2$

However, we don't want to change the value of the expression in question, and therefore- we will also subtract this term from the expression,

That is, we will add and subtract the term (or expression) we need to "complete" to the form of a squared binomial,

In the next calculation, the "trick" is demonstrated (two lines under the term we added and subtracted from the expression),

Then- we will put into the squared binomial form the appropriate expression (demonstrated with colors) and in the last stage we will further simplify the expression:

$(\sqrt{2}x)^2+2\cdot \sqrt{2}x\cdot 7-15\\ (\sqrt{2}x)^2+2\cdot\sqrt{2}x\cdot 7\underline{\underline{+7^2-7^2}}-15\\ (\textcolor{red}{\sqrt{2}x})^2+2\cdot \textcolor{red}{\sqrt{2}x}\cdot \textcolor{green}{7}+\textcolor{green}{7}^2-49-15\\ \downarrow\\ (\textcolor{red}{\sqrt{2}x}+\textcolor{green}{7})^2-49-15\\ \downarrow\\ \boxed{(\sqrt{2}x+7)^2-64}$

Thus- we obtained the completing the square form for the given expression,

Let's summarize the development stages, we will do this now within the given equation:

$2x^2+14\sqrt{2}x-15=0 \\ (\sqrt{2}x)^2+2\cdot \sqrt{2}x\cdot 7-15=0\\ (\textcolor{red}{\sqrt{2}x})^2+2\cdot \textcolor{red}{\sqrt{2}x}\cdot \textcolor{green}{7}\underline{\underline{+\textcolor{green}{7}^2-7^2}}-15=0\\ \downarrow\\ (\textcolor{red}{\sqrt{2}x}+\textcolor{green}{7})^2-49-15=0\\ \downarrow\\ \boxed{(\sqrt{2}x+7)^2-64=0}$

Now, we can isolate from this expression a simpler algebraic expression,

We will do this by transferring sides and extracting a square root:

$(\sqrt{2}x+7)^2-64=0\\ (\sqrt{2}x+7)^2=64\hspace{6pt}\text{/}\sqrt{\hspace{6pt}}\\ \downarrow\\ \boxed{\sqrt{2}x+7=\pm8}$

(We will remember of course that extracting a square root from both sides of the equation involves considering two possibilities - with a positive sign and with a negative sign)

Let's note now that we are interested in the value of the expression:

$\sqrt {2}x+5=\text{?}$

Which we can easily extract from the equations we obtained,

At this stage we will emphasize two important things:

A. We obtained two equations requiring two values with opposite signs for the same expression

$\sqrt{2}x+7=\pm8$

But it's easy to understand that these two equations cannot be held together unless the expression equals 0, which is not the case here.

B. Because of this, we need to separate and solve each one independently in order to get all the possibilities for the value of the requested expression,

We will continue, and refer to each equation separately, first we will try to identify the requested expression, and then isolate it, in each equation separately:

$\sqrt{2}x+7=\pm8 \\ \underline{\textcolor{blue}{\sqrt{2}x+5}}+2=\pm8 \\ \downarrow\\ \sqrt{2}x+5+2=8 \rightarrow\boxed{\sqrt{2}x+5=6} \\ \sqrt{2}x+5+2=-8 \rightarrow\boxed{\sqrt{2}x+5=-10} \\ \downarrow\\ \boxed{\sqrt{2}x+5=6,\hspace{4pt}-10}$

Therefore, the correct answer is answer D.