Solving equation with change of variable

Question

$$x^4+x^3-10x^2+x+1=0$$
I know the whole solution, but i can't understand some parts
So first we divide whole equation by $x^2$, then
$$\begin{eqnarray} x^2+x-10+1/x+1/x^2 &=& 0 \\ \frac{x+1}{x} &=& t \end{eqnarray}$$
So why exactly $\frac{x+1}{x}$ ? How do i need to know what i need to change?
Then $x^2+1/x^2=t^2-2$
Where did the 10 went?
Thanks instantly for help.

Answer 1

Setting

$$x+\frac1x=t$$

we have

$$x^2-10+\frac1{x^2}=\left(x+\frac1x\right)^2-12=t^2-12$$

thus

$$x^2+x-10+\frac1x+\frac1{x^2}=0\implies t^2+t-12=0$$

Answer 2

For all real $k$ we obtain: $$x^4+x^3-10x^2+x+1=\left(x^2+\frac{1}{2}x+k\right)^2-\frac{1}{4}x^2-k^2-kx-2kx^2-10x^2+x+1=$$ $$=\left(x^2+\frac{1}{2}x+k\right)^2-\left(\left(2k+\frac{41}{4}\right)x^2+(k-1)x+k^2-1\right).$$ Thus, for $k=1$ we obtain: $$x^4+x^3-10x^2+x+1=\left(x^2+\frac{1}{2}x+1\right)^2-\frac{49}{4}x^2=$$ $$=(x^2-3x+1)(x^2+4x+1).$$ Now, solve two quadratic equations.

Answer 3

Because the coefficients are palindromic (read the same backwards as forwards), whenever $x=a$ is a solution of the equation, so is $x=\frac 1a$.

Thus the roots will be $x=\pm 1$ (which are easily identified and factored out) or come in pairs which will give factors of the form $$(x-a)(x-\frac 1a)=x^2-(a+\frac 1a)x+1$$If we set this equal to $0$ and rearrange we get $x+\frac 1x=a+\frac 1a$, from which we see that if we substitute $y=x+\frac 1x$ these two factors become a linear factor in $y$. Once you have the value of $y$ it is a simple matter of solving a quadratic to find the roots of the original equation.

"The substitution works" is one justification, but I hope this gives an idea as to why it works.

Answer 4

The idea here is to find some change of variable which makes the equation easier to solve. We might have guesses as to what to change, but we know we were right only if it ends up working out.

Here, you might guess to substitute $t = x + \frac{1}{x}$ because there are four different powers of $x$ in the equation -- $x^2$, $x$, $x^{-1}$, and $x^{-2}$ -- and this substitution lets us combine $x$ and $x^{-1}$ into $t$, and hopefully $x^2$ and $x^{-2}$ into $t^2$, so that we'll only end up with two different powers of $t$, which is much easier to solve.

So let's try it and see how it works out.

From $t = x + \frac{1}{x}$ we compute $t^2 = (x + x^{-1})^2 = x^2 + x^{-2} +2$. So $x^2 + x^{-2} = t^2 - 2$. This is good because $x^2 + x^{-2}$ appears in our equation.

Our main equation is (rewritten to show where we will substitute): $$ (x^2 + x^{-2}) + (x + x^{-1}) - 10 = 0 $$ (The $10$ is still there, we just won't substitute for it.) We now place our substitutions: $$ (t^2 - 2) + (t) - 10 = 0$$ $$ t^2 + t - 12 = 0$$ And now we can proceed to solve for $t$.

Answer 5

Another technique worth considering is that on the real field you can theoretically factor polynomials into polynomials of degree at most $2$.

Can all real polynomials be factored into quadratic and linear factors?

When the coefficients are favourable, like in the present case the coefficients of $x^4$ and $x^0$ are both $1$, we can restrict our search for $$\begin{align}x^4+x^3-10x^2+x+1&=(x^2+ax+1)(x^2+bx+1)\\&=x^4+(a+b)x^3+(ab+2)x^2+(a+b)x+1\end{align}$$

Identifying the coefficients gives $\begin{cases}a+b=1\\ab+2=-10\end{cases}\iff a,b=-3,4$

And we are left with two quadratics we are able to solve.

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Now we will see how this method relates to the suggested substitution $t=x+\frac 1x$

$f(x)=x^4+x^3−10x^2+x+1$

We can notice that $f(\frac 1x)=\frac 1{x^4}+\frac 1{x^3}-\frac{10}{x^2}+\frac 1x+1=\dfrac{f(x)}{x^4}$

So since $f(0)\neq 0$, if $r$ is root of $f(x)=0$ then $\frac 1r$ is a root as well.

It means we can factor

$\begin{align}f(x) &=(x-a)(x-\frac 1a)(x-b)(x-\frac 1b)\\ &=\left(x^2-(a+\frac 1a)x+1\right)\left(x^2-(b+\frac 1b)x+1\right)\\ &=(x^2-t_ax+1)(x^2-t_bx+1) \end{align}$

With the substitution $t_a=a+\frac 1a$ and $t_b=b+\frac 1b$, this is exactly what we solved in the first method.