1. **State the problem:** Solve the equation $e^{x^2} - 2x = 1$ for $x$. 2. **Rewrite the equation:** $$e^{x^2} - 2x = 1$$ Move all terms to one side: $$e^{x^2} - 2x - 1 = 0$$ 3. **Analyze the equation:** This is a transcendental equation involving both an exponential function and a polynomial term. Such equations usually cannot be solved algebraically with elementary functions. 4. **Check for possible solutions by inspection:** Try $x=0$: $$e^{0^2} - 2(0) - 1 = e^0 - 0 - 1 = 1 - 1 = 0$$ So, $x=0$ is a solution. 5. **Check for other solutions:** Rewrite as: $$e^{x^2} = 1 + 2x$$ Since $e^{x^2} > 0$ for all real $x$, the right side must be positive: $$1 + 2x > 0 \\ 2x > -1 \\ x > -\frac{1}{2}$$ 6. **Test values near $x=0$ and $x=-\frac{1}{2}$:** At $x=-0.4$: $$e^{(-0.4)^2} = e^{0.16} \approx 1.1735$$ $$1 + 2(-0.4) = 1 - 0.8 = 0.2$$ Left side $>$ right side. At $x=1$: $$e^{1^2} = e^1 \approx 2.7183$$ $$1 + 2(1) = 3$$ Left side $<$ right side. At $x=1.2$: $$e^{1.44} \approx 4.22$$ $$1 + 2(1.2) = 3.4$$ Left side $>$ right side. So the function crosses the right side between $x=1$ and $x=1.2$. 7. **Use numerical methods (e.g., Newton's method) to approximate the second root:** Initial guess $x_0=1$. 8. **Summary:** - One exact solution is $x=0$. - Another solution exists approximately between $1$ and $1.2$. **Final answer:** $$x=0 \quad \text{and approximately} \quad x \approx 1.1$$