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I've realised that we often find out how to solve equations in terms of $x$ that are quadratics, or cubics, or even quartics, but the other day I came up with the following question, and realised I've never managed to find out how to solve an equation of this sort:

$$x^{1.2} + 2x - 7 = 0$$

Solve for all values of $x$

In this example, there is an instance of $x$ with a decimal exponent (i.e. $1.2$). Is it possible to solve this equation to find both solutions algebraically, without trial and error or iteration? I've tried searching all over the internet and YouTube but I can't seem to find an answer anywhere?

Thank you in advance.

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  • $\begingroup$ You can only solve this equation numerically $\endgroup$ – Andrei Aug 14 '18 at 13:55
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You can transform this equation into a polynomial as follows but in this case, the polynomial is of degree 6 with no rational roots, so it still has to be solved numerically

First, isolate $x^{1.2}$ on one side: $$x^{1.2}+2x-7=0$$ $$x^{1.2}=-2x+7$$ Now we can raise both sides to the fifth power to make all exponents integers: $$\left(x^{1.2}\right)^5=\left(-2x+7\right)^5$$ $$x^6=-32x^5+560x^4-3920x^3+13720x^2-24010x+16807$$ Which can be written in standard form as: $$x^6+32x^5-560x^4+3920x^3-13720x^2+24010x-16807=0$$

Note that in this case, we couldn't solve the polynomial algebraically, but with an equation such as $x^{1.5}+2x-7=0$, you can solve the polynomial algebraically (although it is rather cumbersome since it becomes a cubic equation with no rational roots).

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  • $\begingroup$ You could also use the substitution $x=u^5$, which yields the equation $u^6+2u^5-7=0$. Once you solve for $u$, $x$ is the fifth root of those solutions. $\endgroup$ – InterstellarProbe Aug 14 '18 at 14:05
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An equation with rational exponents can be turned in one with integer exponents, i.e. a polynomial, by a change of variable, using the $\text{lcm}$ of the denominators of the exponents.

E.g.

$$x^{6/5}+2x+7=0\to (y^5)^{6/5}+2y^5+7=y^6+2y^5+7=0.$$

When there are irrational exponents, this is not possible and in principle you have to resort to numerical methods.

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