Evaluate x^2 + 49/x^2 When x + 7/x = 2: No-Solve Method

We want to calculate the value of the expression:

$x^2+\frac{49}{x^2}=\text{?}$

based on the given equation:

$x+\frac{7}{x}=2$

but without solving it for x,

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

in the expression we want to calculate - there are terms in second power only,

therefore we understand that apparently we need to square the expression on 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 emphasize something worth noting that happens in the given mathematical structure in the expression in question (the mathematical structure where a term and its proportional inverse are added):

$x+\frac{7}{x}=2 \hspace{6pt}\text{/}()^2\\ (x+\frac{7}{x})^2=2^2\\ \downarrow\\ x^2+2\cdot \textcolor{blue}{x\cdot \frac{7}{x}}+ \frac{7^2}{x^2}=4\\ \downarrow\\ x^2+2\cdot \textcolor{blue}{7}+ \frac{49}{x^2}=4\\$Let's now notice that the "mixed" term in the square formula ($2ab$) gives us - from squaring the mathematical structure in question - a free number, meaning - it's not dependent on variable x, since it involves multiplication between an expression with a variable and its proportional inverse,

This fact actually allows us to isolate the desired expression from the equation we got and get its value (which is not dependent on the variable) even without knowing the value of the unknown (or unknowns) that solves the equation:

$x^2+2\cdot \textcolor{blue}{7}+ \frac{49}{x^2}=4\\ x^2+14+ \frac{49}{x^2}=4\\ \boxed{x^2+\frac{49}{x^2}=-10}$

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

How many (real) solutions does the given equation have?,

For this, let's examine the equation we got by squaring the given equation and by moving terms between sides,

Let's note that on the left side there's an expression that is a sum of two positive terms, and therefore it is certainly an expression that has only positive values:

$x^2+\frac{49}{x^2}\rightarrow x^2,\hspace{4pt}\frac{49}{x^2}\\ \downarrow\\ x^2>0 \hspace{6pt}(x\neq0)\\ \frac{49}{x^2}>0 \hspace{6pt}(x\neq0)\\ \downarrow\\ \boxed{x^2+\frac{49}{x^2}>0}$

This is because even powers will always give positive results or zero (which in this case is ruled out by the domain of definition of the unknown in the equation), and since the sum of two positive terms is positive too,

We'll continue and examine the right side of the resulting equation, and conclude that this is impossible, since the resulting equation requires that an expression that is certainly positive (on the left side), to be negative (the expression on the right side):

$\textcolor{red}{x^2+\frac{49}{x^2}}>0\leftrightarrow\textcolor{red}{x^2+\frac{49}{x^2}=-10}\\ \updownarrow(\text{but})\\\ \textcolor{red}{-10}<0$$$

Therefore there is no (real) value of the unknown x that when substituted in the equation:

$x^2+\frac{49}{x^2}=-10$will give a true statement.

And certainly any solution to the given equation must satisfy the equation we got by squaring both sides (mentioned above),

Therefore- there is no (real) solution to the given equation.

(And we concluded this without trying to solve it for the unknown x)

Therefore the correct answer is answer C.