Finding roots of $-3x^{1.25}-3x+10$

Question

I'm in a math workshop, where one of the problems given was $y=–3x^{1.25} –3x+10$. Much to my frustration, the only stated way to find roots was finding x by trial and error.

Is there any way to accurately find the roots of this function using a root-finding algorithm – not unlike the quadratic formula?

Answer 1

Write $x=u^4$. Then we have to solve $$ -3u^5-3u^4+10=0$$

A numerical solution $u\approx1.1198125298329972524296485747877379384$ (so $x\approx 1.5724660961391427815788005324759241792$) is readily found, however an exact solution with radicals is ... unlikely. At least for general equations of fifth degree there is (provably!) no such solution method possible. One would need to go into some deeper detail to check if this specific equation also has no solution by radicals. Actually doing this would be overkill for a precalculus course and Thus I won't, but I bet it is the case.

Answer 2

I assume that the question is to find the roots of the equation you've given. That is, we need to find $x$ such that $$ 3x^{1.25} + 3x - 10 = 0 $$ We can rearrange the question to get $$ 3x^{1.25} = 10 - 3x\\ (3x^{1.25})^4 = (10 - 3x)^4\\ 81x^5=81 x^4-1080 x^3+5400 x^2-12000 x+10000\\ 81x^5-81 x^4+1080 x^3 - 5400 x^2 + 12000 x - 10000=0 $$ The question then becomes "what $x$ is the real root of the above equation?" As it ends up, there is no neat solution like the quadratic formula for polynomials of degree $5$ or higher, and this seems to be an example of an irreducible quintic equation. Trial and error does seem to be the best approach after all.

Answer 3

There are algorithms for solving this, but they require calculus. One of the most commonly used methods is Newton's Method, which approaches the root using tangent lines. Let $y=f(x)$ and choose a seed $x_0$. To get $x_{n+1}$ from $x_n$, first compute the tangent line: $$\tilde{y}_n=f(x_n)+f'(x_n)(x-x_n),$$ where $f'$ represents the first derivative of $f$ with respect to $x$. From there, find the $x$-intercept by setting $\tilde{y}_n=0$ and solving for $x$. This intercept is your $x_{n+1}$. In short, $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}.$$ This iterative method converges to the root of the function. For a faster convergence, you can add additional terms from the Taylor polynomial $$f(x)=\sum_{k=0}^\infty \frac{f^{(k)}(x_n)(x-x_n)^k}{k!}.$$