4
$\begingroup$

Today, I had this question on a Maths test about Algebra. This was the equation I had to solve:

$$(1-x)(x-5)^3=x-1$$

I worked away the brackets and subtracted $x-1$ from both sides and was left with this:

$$-x^4+16x^3-90x^2+199x-124=0$$

Problem is, I haven't a clue how to solve this? First thing I tried was replacing $x^2$ with another variable like $u$ but that got me no further. Dividing the whole equation by $x^2$ (as is suggested by a lot of sites on this matter) also did not get me any further. I then tried something incredibly ludicrous;

$$(ax+b)(cx^3+dx^2+ex+f)=0$$ $$ \left\{ \begin{aligned} ac &= -1 \\ ad + bc &= 16 \\ ae + bd &= -90 \\ af + be &= 199 \\ bf &= -124 \end{aligned} \right. $$

which got even worse when there were 3 brackets;

$$(ax+b)(gx+h)(ix^2+jx+k)=0$$ $$ \left\{ \begin{aligned} ac &= agi &&= -1 \\ ad + bc &= agj + ahi + bgi &&= 16 \\ ae + bd &= agk + ahj + bgj + bhi &&= -90 \\ af + be &= ahk + bgk + bhj &&= 199 \\ bf &= bhk &&= -124 \end{aligned} \right. $$

only to be left with no result.

When using Wolfram Alpha on this question, it performs a rather strange step I don't understand:

$$-x^4+16x^3-90x^2+199x-124=0$$ $$\downarrow$$ $$-((x-4)(x-1)(x^2-11x+31))=0$$

Could somebody explain how to properly tackle this problem? And if possible, also show me how to get the non-real answers for it. Thanks.

$\endgroup$
  • 3
    $\begingroup$ You needn't expand: first, observe that $x=1$ is a solution. Now, looking for other solutions (at most 3 other possible ones), you can divide each side by $x-1$ and obtain the equation $(x-5)^3=-1$ to solve (for values $x\neq 1$). $\endgroup$ – Clement C. May 20 '15 at 13:13
8
$\begingroup$

Euh... I think you overcomplicated things here...

$(1-x)(x-5)^3=x-1$ is equivalent to $(1-x)[(x-5)^3+1]=0$

Either $x=1$ or $(x-5)^3=-1$...

$\endgroup$
  • $\begingroup$ I feel so stupid, that I didn't see this in the first place... Thanks for the great answer! $\endgroup$ – Martijn May 20 '15 at 13:19
  • $\begingroup$ @TitoPiezasIII Done :) $\endgroup$ – Martijn May 20 '15 at 13:27
  • $\begingroup$ You can also use the identity $a^3+1=(a+1)(...)$ $\endgroup$ – k1.M May 20 '15 at 14:00
4
$\begingroup$

There are already answers on how to get the real solutions, so I will only show you the non-real solutions.

You have obtained that $(x-5)^3=-1$. Expanding and simplifying we get: $$x^3-15x^2+75x-124=0$$ However, we know that $x=4$ is a solution so we can say that $(x-4)(ax^2+bx+c)=0$. You can equate coefficients or use polynomial division, but as you have already found with Wolfram Alpha, $(ax^2+bx+c)=(x^2-11x+31)$.

To solve, complete the square: $$x^2-11x+31=0$$ $$(x-\frac{11}{2})^2-\frac{121}{4}+31=0$$ $$(x-\frac{11}{2})^2=-\frac{3}{4}$$ $$x-\frac{11}{2}=\sqrt{-\frac{3}{4}}= \pm \frac{\sqrt{3}}{2}i$$ $$x=\frac {11 \pm i \sqrt{3}}{2}$$

$\endgroup$
  • 1
    $\begingroup$ Awesome work! I have two kinds of mathematics in the Netherlands and the other kind is about imaginary numbers a.t.m. so this will definitely come in handy, I'm sure! $\endgroup$ – Martijn May 20 '15 at 13:45
  • $\begingroup$ In general the solutions to $(x-a)^m=-1$ are $x_n=e^{i\pi\frac{2n-1}{m}}+a$, for $n=1,2,\dots m$. $\endgroup$ – Kwin van der Veen May 21 '15 at 4:10
2
$\begingroup$

From the beginning: $$(1-x)(x-5)^3=x-1\\ (1-x)(x-5)^3+1-x=0\\ (1-x)(x-5)^3+(1-x)=0\\ (1-x)[(x-5)^3+1]=0\\$$

This implies $1-x=0$ or $(x-5)^3=-1$. I believe you can solve these.

$\endgroup$
  • 4
    $\begingroup$ Isn't there a sign that got flipped between the first and second line? $\endgroup$ – Clement C. May 20 '15 at 13:18
  • $\begingroup$ Sorry fixed it. @ClementC. $\endgroup$ – KittyL May 20 '15 at 13:24
2
$\begingroup$

I know that the problem has already been answered but I want to show you a more general method, let's suppose that you don't se how to rewrite the equation:

$-x^4+16x^3-90x^2+199x-124=0$

Or I prefer to write:

$x^4-16x^3+90x^2-199x+124=0$

You can use something called the Ruffini rule: search for integers divisor (both positive and negative) of the constant term and then set $x$ equals to the them and see if one of them is a solution. Starting from one we have:

$1-16+90-199+124=0$

So $x=1$ is a solution, now via Ruffini's rule, which can be seen here, we can rewrite the equation as:

$(x-1)(x^3-15x^2+75x-124)=0$

Now you have $3$ options to end this exercise:

$1.)$ Note that the second factor is a perfect cube;

$2.)$ Use Ruffini's rule again;

$3.)$ Use the general formula for third degree equation (which I'd not advise you to if your interested only in real solution).

This is a more general method so I hope this will help you in the future!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.