How to solve $e^{2x - 1} = -x^2 + 3x - 0.25$

Question

I know the answer is $x=0.5$, but I only know it because my calculator showed me the intersection point of the two graphs. What I want to know is how to get to this result algebraically. I've tried to use the Lambert-W function, but I ended up with $$W((1-2x)/(-x^2+3x-0.25))=1-2x$$ and this didn't help me either. If smn has an idea on how to solve this, pls let me know. Obs: I don't know how to use the Newton-raphson's method, so that's why I haven't tried it.

Answer 1

Consider the two curves $y=e^{2x-1}$ and the parabola $y=-x^2+3x-0.25$

They have the same value $y=1$ at $x=0.5$

Furthermore $y'=2e^{2x-1}$ and $y'(0.5)=2$ so the tangent line at $(0.5,1)$ has equation $y-1=2(x-0.5)\to y=2x$

For the parabola $y'=-2x+3$ and $y'(0.5)=2$

the tangent at $(0.5,1)$ is the same than the exponential curve.

The curves are tangent at $(0.5,1)$

$y''=4e^{2x-1}$ is positive at $x=0.5$ while for the parabola $y''=-2$ is negative for any $x$ so the parabola has convexity down (is less than the tangent in any $x$), while the exponential has convexity up (is more than the tangent at any $x$)

So $x=0.5$ is the one and only solution of the equation

Hope it helps

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Answer 2

For $$e^{2x-1} + x^2 = 3 x - \frac{1}{4}$$ it can be seen that \begin{align} e^{2x-1} + x^2 &= 3 x - \frac{1}{4} \\ e^{2x -1} + x^{2} - x + \frac{1}{4} &= 2 x = 2 \left(x - \frac{1}{2}\right) + 1 \\ e^{2 x -1} + \left(x - \frac{1}{2}\right)^{2} - 2 \left( x - \frac{1}{2} \right) + 1 &= 2 \\ e^{2 x - 1} + \left( \left(x - \frac{1}{2}\right) - 1 \right)^{2} &= 2 \\ e^{2 \left(x - \frac{1}{2}\right)} + \left(x - \frac{3}{2}\right)^{2} &= 2. \end{align} The only solution to this equation is $$x = \frac{1}{2}.$$