MHB Solve Equation: $x^4+2x^3-x^2-6x-3=0$

[solakis1]

Solve the following equation:

$x^4+2x^3-x^2-6x-3=0$

 

[Opalg]

Brute force solution. :censored:

Spoiler

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The graph of the function $x^4 + 2x^3 - x^2 - 6x - 3$ shows that it has two real roots, approximately 1.62 and -0.62, with sum close to 1 and product close to -1.So it looks as though they might be the roots of the quadratic $x^2 - x - 1$. It is then easy to factorise the function as $$x^4 + 2x^3 - x^2 - 6x - 3 = (x^2 - x - 1)(x^2 + 3x + 3).$$ So the roots are $x = \frac12\bigl(1\pm\sqrt5\bigr)$ and $x = \frac12\bigl(-3\pm i\sqrt3\bigr)$.

That factorisation can be written as $$\begin{aligned}x^4 + 2x^3 - x^2 - 6x - 3 &= \bigl((x^2+x+1) - 2(x+1)\bigr)\bigl((x^2+x+1) + 2(x+1)\bigr)\\ &= (x^2+x+1)^2 - 4(x+1)^2.\end{aligned}$$ Maybe a more elegant solution could be found by approaching the problem from that direction, expressing $x^4 + 2x^3 - x^2 - 6x - 3$ as the difference of two squares?

 

[I like Serena]

Another brute force solution based on an educated guess.

Spoiler

Let's try to write it as $(x^2+ax+c)(x^2+bx+d)$.
We'll assume that $c$ and $d$ are integers, which means that $c=\pm 3$ and $d=\mp 1$ or vice versa. Due to symmetry, we can pick $c$ and $d$ as written without losing a solution.
So we have either $(x^2+ax+3)(x^2+bx-1)$ or $(x^2+ax-3)(x^2+bx+1)$.

For the first case, we have:
$$(x^2+ax+3)(x^2+bx-1)=x^4+(a+b)x^3+(3-1+ab)x^2+(-a+3b)x-3$$ which must be equal to $$x^4+2x^3-x^2-6x-3$$
It follows that:
$$\begin{cases}a+b=2 \\ 3-1+ab=-1 \\ -a+3b=-6 \end{cases} \implies \begin{cases} a=3 \\b=-1 \end{cases}$$
That means that we've found $(x^2+3x+3)(x^2-x-1)=0$. It also means that we don't have to analyze the other case, since this is good enough.

We can now solve it with the usual quadratic formula to find $x=\frac 12(-3\pm i{\sqrt 3})$ and $x=\frac 12(1\pm \sqrt 5)$.

 

[solakis1]

Another brute force solution based on an educated guess.

Spoiler

Let's try to write it as $(x^2+ax+c)(x^2+bx+d)$.
We'll assume that $c$ and $d$ are integers, which means that $c=\pm 3$ and $d=\mp 1$ or vice versa. Due to symmetry, we can pick $c$ and $d$ as written without losing a solution.
So we have either $(x^2+ax+3)(x^2+bx-1)$ or $(x^2+ax-3)(x^2+bx+1)$.

For the first case, we have:
$$(x^2+ax+3)(x^2+bx-1)=x^4+(a+b)x^3+(3-1+ab)x^2+(-a+3b)x-3$$ which must be equal to $$x^4+2x^3-x^2-6x-3$$
It follows that:
$$\begin{cases}a+b=2 \\ 3-1+ab=-1 \\ -a+3b=-6 \end{cases} \implies \begin{cases} a=3 \\b=-1 \end{cases}$$
That means that we've found $(x^2+3x+3)(x^2-x-1)=0$. It also means that we don't have to analyze the other case, since this is good enough.

We can now solve it with the usual quadratic formula to find $x=\frac 12(-3\pm i{\sqrt 3})$ and $x=\frac 12(1\pm \sqrt 5)$.

wait a minite if you substitute the 1st equation into the3rd don't you get b=2

 

[solakis1]

You

Brute force solution. :censored:

Spoiler

[DESMOS]{"version":7,"graph":{"viewport":{"xmin":-10,"ymin":-8.406231118816823,"xmax":10,"ymax":8.406231118816823}},"randomSeed":"fb3a103f1f6795c6aa824f4360e302c0","expressions":{"list":[{"type":"expression","id":"1","color":"#c74440","latex":"y=\\ x^{4}\\ +\\ 2x^{3}\\ -\\ x^{2}\\ -\\ 6x\\ -\\ 3"}]}}[/DESMOS]
The graph of the function $x^4 + 2x^3 - x^2 - 6x - 3$ shows that it has two real roots, approximately 1.62 and -0.62, with sum close to 1 and product close to -1.So it looks as though they might be the roots of the quadratic $x^2 - x - 1$. It is then easy to factorise the function as $$x^4 + 2x^3 - x^2 - 6x - 3 = (x^2 - x - 1)(x^2 + 3x + 3).$$ So the roots are $x = \frac12\bigl(1\pm\sqrt5\bigr)$ and $x = \frac12\bigl(-3\pm i\sqrt3\bigr)$.

That factorisation can be written as $$\begin{aligned}x^4 + 2x^3 - x^2 - 6x - 3 &= \bigl((x^2+x+1) - 2(x+1)\bigr)\bigl((x^2+x+1) + 2(x+1)\bigr)\\ &= (x^2+x+1)^2 - 4(x+1)^2.\end{aligned}$$ Maybe a more elegant solution could be found by approaching the problem from that direction, expressing $x^4 + 2x^3 - x^2 - 6x - 3$ as the difference of two squares?

And if you divide $x^4+2x^3-x^2-6x-3$ by $x^2-x-1 $ the result will be $x^2+3x+3 $
And you do not need that brutal factorization;)

 

[solakis1]

Solve the following equation:

$x^4+2x^3-x^2-6x-3=0$

AN easy solution:

$x^4+2x^3-x^2-6x-3=x^4+2x^3+2x^2-3x^2-6x-3=x^4+2x^2(x+1)-3(x+1)^2=
(\frac{x^2}{x+1)})^2+2\frac{x^2}{x+1}-3=0$
devide by $(x+1)^2$
The solutions of this quadratic equation are:
because if you put $y=\frac{x^2}{x+1}$ you get the quadratic equation $y^2+2y-3=0$

$\frac{x^2}{x+1}=1$ .............1
or
$\frac{x^2}{x+1}=-3$.............2

(1) gives the wanted solution
(2) gives no real solutions
Note : $x+1\neq 0$

sorry spoiler does not work

 

[I like Serena]

wait a minite if you substitute the 1st equation into the3rd don't you get b=2

If we add $a+b=2$ and $-a+3b=-6$ together, we get $4b=-4 \implies b=-1$.

 

[solakis1]

If we add $a+b=2$ and $-a+3b=-6$ together, we get $4b=-4 \implies b=-1$.

My GOD i did this stupid substitution 3 times