Animated Solution for Mathematics - Quadratic Equations: The number of real solutions of the equation $3\left(x^2+\frac{1}{x^2}\right)-2\left(x+\frac{1}{x}\right)+5=0$, is

My dear student, welcome to a beautiful exploration of algebraic symmetry. When you first look at the equation

it might seem like a daunting, high-degree polynomial. But look closer. Do you see the hidden harmony?

The expression is perfectly symmetric with respect to $x$ and $\frac{1}{x}$. This is not a coincidence; it is an invitation to simplify.

To tame this beast, we introduce a new variable. Let $t=x+\frac{1}{x}$. This is our bridge between the complex-looking original equation and a much friendlier quadratic.

But we cannot simply swap variables without accounting for the squared terms. If we square our substitution, we get

which expands to $x^2+\frac{1}{x^2}+2=t^2$. Therefore, we can elegantly replace $x^2+\frac{1}{x^2}$ with $t^2-2$. This is the key that unlocks the door.

Now, let us substitute these into our original equation:

Expanding this, we get $3t^2-6-2t+5=0$, which simplifies beautifully to

This is a standard quadratic equation. Factorizing it, we split the middle term: $3t^2-3t+t-1=0$, leading us to $(3t+1)(t-1)=0$. Our potential values for $t$ are $t=1$ and $t=-\frac{1}{3}$.

Here is where many brilliant students stumble. We have found values for $t$, but are they valid for real $x$? Recall our substitution $t=x+\frac{1}{x}$.

For any real number $x$, the function $f(x)=x+\frac{1}{x}$ has a very specific range. By the AM-GM inequality, if $x>0$, then $x+\frac{1}{x}\ge 2$. If $x<0$, then $x+\frac{1}{x}\le -2$.

This means that for any real $x$, the magnitude of $t$ must satisfy $|t|\ge 2$.

Look at our values: $t=1$ and $t=-\frac{1}{3}$. Both of these values lie strictly between $-2$ and $2$.

They fall right into the 'forbidden zone' where no real $x$ can exist. Because neither value of $t$ satisfies the condition $|t|\ge 2$, there are no real values of $x$ that can satisfy the original equation.

The number of real solutions is exactly 0. Always remember, in the JEE, the algebra is only half the battle; the constraints are where the true test lies.