How many zeros does a radical equation (eg, $X^{4/3}-5X^{2/3}+6=0$) have?

Question

I want to know if there is a general rule that will give me the answer. I'm not talking about crazy expressions under a radical, just simple variables raised to a fractional exponent like:

$$X^{4/3}-5X^{2/3}+6=0$$

Answer 1

For a case like this we can define $y=x^{\frac 23}$ and make the equation $y^2-5y+6=0$ which has two roots in $y$. Then you can have either square root, so there will be four zeros. You need to carefully specify the range of equations that is acceptable to get any simple rule.

Answer 2

There is no "general rule" for solving all such equations. But your equation is a quadratic equation in disguise. Such questions mostly appear in pre-calculus exercises.

A rule of thumb is that you can make your equation easier by introducing new variables.

Since $x^{m/n}=(x^{1/n})^m$, you can write $$ (x^{4/3})=(x^{1/3})^4,\quad (x^{1/3})^2 $$ so that your equation becomes $$ (x^{1/3})^4-5(x^{1/3})^2+6=0\ \tag{1} $$

Now if you introduce a new variable $y=x^{1/3}$, your equation becomes $$ y^4-5y^2+6=0\tag{2} $$ which looks simpler since there is no "radical" (fraction powers) anymore.

But this is still not easy: it is of power $4$.

But if you observe that $y^4=(y^2)^2$, (2) is the same as $$ (y^2)^2-5y^2+6=0\tag{3} $$ where if you introduce another new variable $w=y^2$, you have a quadratic equation $$ w^2-5w+6=0 $$ which can be solved easily since $w^2-5w+6=(w-2)(w-3)$.

Answer 3

Say that in your equation, the exponents of your unknown $x$ are fractions $$\frac{m_1}{n_1},\ldots,\frac{m_k}{n_k}.$$

Let $N$ be the least common multiple of $n_1,\ldots,n_k$, and let $N = n_ic_i$. Then $$\frac{m_i}{n_i} = \frac{m_ic_i}{n_ic_i} = \frac{m_ic_i}{N}.$$ That means that you can rewrite your equation as an equation in which every radical is an $N$th root. Letting $y=x^{1/N}$, you can then turn your equation into an equation in $y$ which has no radicals, just integral powers. That equation will be an equation of degree $M=\max\{c_1m_1,\ldots, c_km_k\}$, and so has at most $M$ solutions for $y$. When $N$ is odd, each solution for $y$ will give you a unique solution for $x$, yielding at most $M$ solutions for $x$.

If $N$ is even then at least one of the $n_i$ is even, which means that your $x$ must be nonnegative. So nonnegative values of $y$ will give one corresponding solution, but negative values of $y$ will not correspond to solutions of the original equation. In any case, there is at most one $x$ value per $y$-value, so again you get at most $M$ solutions for $x$.