Solve Dis Riddles

?How do I solve 4x2−3x=9

This equation is transcendental, and it would seem that its solutions, other than 

x=3$x=3$, cannot be written in terms of commonly used functions.

With a little analysis, we can determine how many solutions it must have. First, note that limx→±∞f(x)=∓∞${\displaystyle \underset{x\to \pm \mathrm{\infty }}{lim}f(x)=\mp \mathrm{\infty }}$. As the function is continuous, we know there must be at least one zero.

Differentiating f(x)=4x2−3x−9$f(x)=4x^2-3^x-9$ gives f′(x)=8x−ln3exln3$f^{\prime }(x)=8x-\ln 3e^{x\ln 3}$.

We can set this equal to zero to find the number of local extrema.

f′(x)=8x−ln3exln3=0 ⟹$f^{\prime }(x)=8x-\ln 3e^{x\ln 3}=0⟹$

−xln3e−xln3=−(ln3)28$-x\ln 3e^{-x\ln 3}=-\frac{(\ln 3{)}^2}{8}$

We use the Lambert W function to conclude that

x=−1ln3W(−(ln3)28)$x=-\frac{1}{\ln 3}W(-\frac{(\ln 3{)}^2}{8})$

Note that −1e<−(ln3)28<0$-\frac{1}{e}<-\frac{(\ln 3{)}^2}{8}<0$ and for all values of −1e<x<0$-\frac{1}{e}<x<0$, we know that W(x)$W(x)$has two real values. So we conclude that there are two local critical points in the original function. It isn’t hard to show that one is a local maximum and the other a local minimum. (Taking a second derivative immediately shows that there is exactly one inflection point.) So the function can have at most three zeros.

So we know that we seek either 1$1$, 2$2$, or 3$3$ solutions. By inspection we see that x=3$x=3$ is a solution, and this solution occurs for a value of x$x$ that is larger than the local maximum, so we know that the value of f(x)$f(x)$ at the local maximum must be positive. Furthermore, we see that f(0)<0$f(0)<0$. By continuity, this implies that there must be a second solution between zero and the local maximum (which is at a value less than three.) Finally, we see that since f(0)<0$f(0)<0$ and the function grows without bound as x$x$decreases (and thus must eventually become positive for some negative value of x$x$), then again by continuity, we see that there must be a third solution for some negative value of x$x$.

So we know that there are three solutions. One is x=3$x=3$, one is in the interval (0,3)$(0,3)$, and the other is negative.

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With the analysis done, we are left to numerically approximate the solutions as were’re unlikely to be able to write them using any commonly defined function.

There are many ways to do the numerical approximation. My favorite (although certainly NOT the most efficient) is the fixed point iteration method.

We seek a function which has the solution of our equation as a fixed point. Recall that x$x$ is a fixed point of function g$g$ if x=g(x)$x=g(x)$. There are many functions we can find for which the solutions we seek are fixed points.

Two such functions are g±(x)=±3x+9√2$g_{\pm }(x)=\pm \frac{\sqrt{3^x+9}}{2}$. Clearly, the negative solution is a fixed point of g−(x)$g_{-}(x)$ while the positive solutions are fixed points of g+(x)$g_{+}(x)$.

Once we have our functions in mind, we form sequences with x0$x_0$ chosen by us and all future values of xn$x_n$ given by the recursion xn+1=g±(xn)$x_{n+1}=g_{\pm }(x_n)$. Notice that IF these sequences have limits (which I’ll just call x$x$) then those limits must satisfy x=g±(x)$x=g_{\pm }(x)$. Hence, these limits would be fixed points and therefore solutions to our original equation.

To find the positive solution that is between zero and three, we pick an initial guess x0=2$x_0=2$. (We’ll get the same limit for any choice between zero and three, so it doesn’t matter much which we start with.) We choose g+(x)$g_{+}(x)$ to form our sequence. The computer can find many terms in the sequence very quickly. The convergence is fairly slow. (I never promised the most efficient method!) It takes 144 terms to reach the precision of my machine. The limit (to quite a few decimals) is:

x≈2.38136170580652$x\approx 2.38136170580652$

We can take the same approach to find the negative solution. Now we use g−(x)$g_{-}(x)$ to form our sequence, and we need to pick a negative starting value. I’ll choose x0=−1$x_0=-1$(although again, the approach is quite robust to the choice of starting value). Interestingly the convergence is MUCH faster in this case. It takes only 8 terms to converge to the limit up to the accuracy of my machine. We learn that the approximate value of the negative solution is:

x≈−1.51568160861330$x\approx -1.51568160861330$

So the three solutions are:

x1≈−1.51568160861330$x_1\approx -1.51568160861330$

x2≈2.38136170580652$x_2\approx 2.38136170580652$

x3=3