Evaluate x-3/4x=1: Finding Value of c in Related Equation

We want to calculate the value of c in the equation:

$16x^2+\frac{9}{x^2}=c$

based on the given equation:

$x-\frac{3}{4x}=1$

but without solving it for x,

For this, let's first notice that while the given equation deals with terms to the first power only,

in the expression we want to calculate:

$16x^2+\frac{9}{x^2}=\text{?}$

there are terms to the second power only,

therefore we understand that apparently we need to square the left side of the given equation,

We'll remember of course the formula for squaring a binomial:

$(a\pm b)^2=a^2\pm2ab+b^2$

and we'll square both sides of the given equation, later we'll highlight something worth noting that happens in the given mathematical structure in the expression in question (the mathematical structure where a term and its reciprocal are added):

$x-\frac{3}{4x}=1 \hspace{6pt}\text{/}()^2\\ (x-\frac{3}{4x})^2=1^2\\ \downarrow\\ x^2-2\cdot \textcolor{blue}{x\cdot\frac{3}{4x}}+ \frac{3^2}{(4x)^2}=1\\ \downarrow\\ x^2-2\cdot \textcolor{blue}{\frac{3}{4}}+ \frac{9}{16x^2}=1\\$Let's now notice that the "mixed" term in the binomial square formula ($2ab$) gives us - from squaring the mathematical structure in question - a free number, meaning- one that doesn't depend on the variable x, since it involves multiplication between an expression with a variable and its reciprocal,

This fact actually allows us to isolate the desired expression (or close to it) from the equation we get and find its value (which doesn't depend on the variable) even without knowing the value of the unknown (or unknowns) that solves the equation:

$x^2-2\cdot \textcolor{blue}{\frac{3}{4}}+ \frac{9}{16x^2}=1\\ x^2-\frac{6}{4}+ \frac{9}{16x^2}=1\\ x^2-\frac{3}{2}+ \frac{9}{16x^2}=1\\ \boxed{x^2+ \frac{9}{16x^2}=\frac{5}{2}}$

Let's continue and focus on our goal, finding c in the equation:

$16x^2+\frac{9}{x^2}=c$

Note that in order to get the left side in the equation mentioned from the left side in the equation we reached earlier, we'll need to multiply both sides of the equation by 16:

$x^2+ \frac{9}{16x^2}=\frac{5}{2}\hspace{6pt}\text{/}\cdot16\\ 16x^2+\frac{\not{16}\cdot9}{\not{16}x^2}=16\cdot\frac{5}{2}\\ 16x^2+\frac{9}{x^2}=40\\ \updownarrow\\ 16x^2+\frac{\cdot9}{x^2}=c\leftrightarrow c=\text{?}\\ \downarrow\\ \boxed{c=40}$

Now, let's try to answer the additional question asked:

Is every solution of the second equation (with c) also a solution to the given equation?

In other words-

Is each one of the solutions to the equation:

$16x^2+\frac{\cdot9}{x^2}=40$

also a solution to the given equation:

$x-\frac{3}{4x}=1$?

Let's note that we reached the first equation mentioned here from the second equation mentioned (which is the given equation) by squaring both sides of the given equation and moving terms,

Generally- equations that are derived from one another through moving terms, multiplying or dividing by a constant are equivalent, meaning their solutions are identical,

However- equations derived from one another by raising to an even power are not necessarily equivalent, because raising to a square (or any even power), might, since it always yields a non-negative result, add solutions to the equation.

Therefore, we cannot determine (without solving the equation for the unknown) whether every solution to the equation:

$16x^2+\frac{\cdot9}{x^2}=40$

is necessarily also a solution to the equation:

$x-\frac{3}{4x}=1$

and therefore the correct answer is answer B.