How to solve $(a-x)^{1.4}-bx=0$?

I'm trying to solve the following equation algebraically : $$(1.722-x)^{1.4}-0.565x=0$$

I can find the solution using Matlab using symbolic function solve or an approximation using Taylor series- the answer is $1.039.$

I have thought of starting from this one $$(1.722-x)^7-(.565x)^5=0$$ but it turns out that there is no algebraic solutions of polynomial equations of degree $\ge5$ by the Abel–Ruffini Theorem.

Is there a way of doing this ? How ? If not, how do I know ?

Another sub-question that came to me while doing my searches is:

Why is there only one solution provided by Matlab whereas $$(1.722-x)^7-(.565x)^5=0$$ should have $7$ solutions as the equation is of degree $7$?

What about the Abel–Ruffini theorem with an equation of the form $ax^6+bx^3+c$ ? I can solve it using $y=x^3$ and therefore contradicts the Theorem. I suppose I have misinterpreted it.

• Do you have reason to believe it can be solved algebraically? – Gregory Grant Dec 25 '15 at 16:22
• The Abel theorem states that there is no general algebraic formula that works for all polynomials of a certain degree higher than four. It does not say that there are not solutions for specific instances or classes. For instance you demonstrated that through substitution you can use the quadratic equation for degree six polynomials of a certain form. However your approach can't work for all degree six polynomials by the theorem. – Joel Dec 25 '15 at 16:24
• As to the question: "Why is there only one solution provided by Matlab...?" Maple gives 7 values, 1.039... and 3 pairs of complex values. – Bernard Masse Dec 25 '15 at 17:05
• You don't say how you're finding solutions with Matlab, but I'm guessing that you're using a numeric method like fzero or fsolve rather than a symbolic method like solve or vpasolve. – horchler Dec 25 '15 at 18:26
• It is maybe better to put $x^{10/14}+\alpha x=\beta$ where $\alpha=b^{-10/14}$ and $\beta=a\alpha$ – Piquito Dec 25 '15 at 20:39

$$(1722 - 1000x)^7 - 10^6 (565x)^5 = 0$$

Or:

$$2\cdot(861-500x)^7 - 5^7\cdot 113\cdot x^5 = 0$$

The way to determine if this has a solution in radicals is to calculate the Galois group, which is a subgroup of $S_7$, and determine if it is solvable. If so, then there is a solution in radicals.

You can do this calculation with the online calculator for Magma:

P<x>:=PolynomialRing(Rationals());

GaloisGroup(2*(861-500*x)^7 - 5^7 * 113 * x^5);

This returns a group of order $5040$. In other words, the Galois group is exactly $S_7$, and so there is no solution in radicals. There may be solutions in terms of other functions, but this is a question of much broader scope.

Incidentally, there are indeed seven distinct roots of this equation, but six of them are complex.

To begin with, your first two equations cannot be solved algebraically.

Abel–Ruffini Theorem simply states we cannot find radical solutions for the general polynomial of degree 5 or higher.

But we can obviously solve your last problem with simple substitution. Or consider the following:$$x^n=0,n\ge5$$$$x=0$$

The theorem has nothing to hold you back from finding solutions, it simply holds you back from solving $$ax^5+bx^4+cx^3+dx^2+ex+f$$

The failure of Matlab to find all 7 solutions to your polynomial is not without reason. Finding such roots is of high difficulty and methods sometimes fail.

I also note that the other 6 roots can be found here, I consider Wolfram|Alpha to be very reliable.

If you can't see all 6 complex roots, just hit the "show more roots" button.

If you are really interested in such solutions, you could try here. Scroll down for the solution for $x$ and hit that exact form button.

• Abel-Ruffini tells us literally nothing about whether a specific polynomial equation can be solved algebraically. – Wojowu Dec 26 '15 at 18:08
• @Wojowu Exactly. That's the mistake the OP made about the theorem. – Simply Beautiful Art Dec 26 '15 at 18:12

$a=1.722-x$ -------------TO MAKE LIFE EASIER

$(1.722-x)^{1.4} - .565x=0$ -------here we substitute (1.722-x) with a

$a^{1.4}+.565a-.97293=0$

$a^7+(.565a^5)=(.97293)^5$ ------multiply the powers by 5 to have an integer.

$\ln(a^7)+\ln(.565)+ln(a^5)=ln(.97293^5)$ --- Here, we use natural logarithm-

also note that ln(.565*a^5) = ln(.565)+ ln(a^5)

$\ln(a^7)+\ln(.565)+ln(a^5)=ln(.97293^5)$ ---------used the top concept

$\ln(a^7)+\ln(a^5)=ln(.97293^5)-ln(.565)$----------Put together like terms.

$\ln(a^7)+\ln(a^5)= 4337/10000$ ----------ln(.97293^5)-ln(.565)= ~.4337

$70000\ln(a)+50000\ln(a)=4337 --------------\ln(a^7) = 7\ln(a)$

$12000\ln(a)=4337$

$12\ln(a)=4337/10000 -------- \ln(a^12)=\ln(e^{4337/10000})$

$a^12=e^{4337/10000} -------- \ln(e^a)=a$

$a=e^{4337/120000}$

$a=1.0368$

This is the value with no negative or imaginary solution.

If you have any questions, private message me.

• A lot of your "algebra" doesn't make sense to me. Could you explain every line please? – Bernard Masse Dec 26 '15 at 1:57
• Wow look at that, Simple Art made it look like gold. Can you pin point Bernard Masse, which part you don't understand? I think now it looks clear and simple. – EyoelD Dec 26 '15 at 2:18
• I tried my best to make it clear but for some reason the editing doesn't look that pretty as Simple Art made it. – EyoelD Dec 26 '15 at 2:34
• There's a major error going from line 4 to line 5: $a+b=c$ does not imply $\ln a + \ln b = \ln c$. I see a couple of minor mistakes here and there as well. – epimorphic Dec 26 '15 at 2:43
• In the 5th line, it appears that you are stating that the logarithm of a sum is a sum of the logarithms. – Michael Burr Dec 26 '15 at 2:43