SS E - The number of real solutions of the equation e** + 4e3* - 58e2* + 4X +1 = 0 is P 2a +

1. Recognising Exponential–Polynomial Mix

The equation $e^{4x} + 4e^{3x} - 58e^{2x} + 4e^{x} + 1 = 0$ contains powers of $e^{x}$. Because $e^{x}$ is always positive, we can safely set $t = e^{x} \quad (t>0).$ This converts the exponential equation to a pure polynomial:

$t^{4} + 4t^{3} - 58t^{2} + 4t + 1 = 0.$

2. Spotting Palindromic Structure

A polynomial is called palindromic when its coefficients read the same forward and backward. Dividing by $t^{2}$ (which is allowed since $t>0$) gives a symmetric expression in $t$ and $\dfrac1t$:

$t^{2} + 4t - 58 + 4\frac1t + \frac1{t^{2}} = 0.$

Group the terms via the identity $t^{2} + \frac1{t^{2}} = \left(t + \frac1t\right)^2 - 2.$ Define $u = t + \frac1t.$

Then the equation reduces to a quadratic in $u$:

$u^{2} + 4u - 60 = 0.$

3. Solving the Quadratic in $u$

Solve using the quadratic formula:

What is the question?

We are asked how many real‐number answers (roots) satisfy the funny looking equation made with the exponential function $e^x$:

$e^{4x} + 4e^{3x} - 58e^{2x} + 4e^{x} + 1 = 0$

Idea in kid language

1. Change the scary thing. Every time we see $e^{x}$ we put a new simpler letter $t$ (because $e^{x}$ is always positive).
2. The giant equation becomes a simpler fourth‐degree (quartic) rule in $t$.
3. Notice the numbers in that new rule read the same forwards and backwards (palindromic). That special symmetry lets us shrink it to a second‐degree (quadratic) rule in $t + \dfrac1t$.
4. Solve the small quadratic, check which answers are possible (remember $t>0$), then turn them back to $x$ using $x = \ln t$.
5. Count how many $x$ values come out.

When you do this carefully you get exactly two real answers.

Step-by-Step Solution

1. Substitution
   $t = e^{x} \quad (t>0)$ gives $t^{4} + 4t^{3} - 58t^{2} + 4t + 1 = 0.$

2. Divide by $t^{2}$ (allowed since $t>0$): $t^{2} + 4t - 58 + 4t^{-1} + t^{-2} = 0.$

3. Use symmetry with $u = t + t^{-1}$:

   $t^{2} + t^{-2} = (t + t^{-1})^{2} - 2 = u^{2} - 2.$

   Plug in: $(u^{2} - 2) + 4u - 58 = 0 \;\Longrightarrow\; u^{2} + 4u - 60 = 0.$

4. Solve the quadratic: $u = \frac{-4 \pm \sqrt{16 + 240}}{2} = \frac{-4 \pm 16}{2}.$ $u_1 = 6, \qquad u_2 = -10.$

5. Feasibility ($t>0 \Rightarrow u \ge 2$): Reject $u=-10$. Keep $u=6$.

6. Recover $t$ from $u=6$: $t + \frac1t = 6 \;\Longrightarrow\; t^{2} - 6t + 1 = 0.$ $t = 3 \pm 2\sqrt2.$ Both $t$ values are positive.

7. Return to $x$ using $x = \ln t$:

   $x_1 = \ln\!(3 + 2\sqrt2), \qquad x_2 = \ln\!(3 - 2\sqrt2).$

8. Count roots: Two real $x$ values.

Answer: Number of real solutions $= \boxed{2}.$