Question

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

• A. 4
• B. 3
• C. 2
• D. 0

Answer 1

The Correct Option is D

The correct answer is (D) : 0
$3\left(x^2+\frac{1}{x^2}\right)-2\left(x+\frac{1}{x}\right)+5=0$
$3\left[{\left(x+\frac{1}{x}\right)}^2-2\right]-2\left(x+\frac{1}{x}\right)+5=0$
Let $x+\frac{1}{x}=t$
$3t^2-2t-1=0$
$3t^2-3t+t-1=0$
$3t(t-1)+1(t-1)=0$
$(t-1)(3t+1)=0$
$t=1,-\frac{1}{3}$
$x+\frac{1}{x}=1,-\frac{1}{3}\Rightarrow$ No solution.

Answer 2

Step 1: Given equation 

The given equation is:

\[ 3\left( x^2 + \frac{1}{x^2} \right) - 2\left( x + \frac{1}{x} \right) = 0 \] 

Step 2: Substitution

We will simplify the equation using the substitution \( x + \frac{1}{x} = t \):

\[ x^2 + \frac{1}{x^2} = t^2 - 2 \] Substituting this into the equation, we get: \[ 3(t^2 - 2) + t + 5 = 0 \] Simplify: \[ 3t^2 - 6 + t + 5 = 0 \quad \Rightarrow \quad 3t^2 - 2t - 1 = 0 \] 

Step 3: Factorizing \p>The equation becomes:

 

\[ 3t^2 - 3t + t - 1 = 0 \] Factorize: \[ (t - 1)(3t + 1) = 0 \] 

Step 4: Solving for t The solutions for \( t \) are: \[ t = 1 \quad \text{or} \quad t = -\frac{1}{3} \] 

Step 5: Checking for real solutions

Now, we substitute \( t = x + \frac{1}{x} \) back to check for real solutions:

- If \( t = 1 \), then \( x + \frac{1}{x} = 1 \). This has no real solution for \( x \). - If \( t = -\frac{1}{3} \), then \( x + \frac{1}{x} = -\frac{1}{3} \). This also has no real solution for \( x \). 

Conclusion:

Thus, there are no real solutions.