- #1

solakis1

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Solve the following equation:

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

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

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In summary, the conversation is about solving the equation $x^4+2x^3-x^2-6x-3=0$ using brute force and a quadratic equation. It is suggested to divide the equation by $x^2-x-1$ and use a substitution to find the solutions. The conversation also includes an error in the solution process and a frustration with the mistake.

- #1

solakis1

- 422

- 0

Solve the following equation:

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

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

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- #2

Opalg

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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?

- #3

I like Serena

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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 $x=\frac 12(1\pm \sqrt 5)$.

- #4

solakis1

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wait a minite if you substitute the 1st equation into the3rd don't you get b=2Klaas van Aarsen said:

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 $x=\frac 12(1\pm \sqrt 5)$.

- #5

solakis1

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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 $Opalg said:

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 you do not need that brutal factorization;)

- #6

solakis1

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AN easy solution:solakis said:Solve the following equation:

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

$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

Last edited:

- #7

I like Serena

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If we add $a+b=2$ and $-a+3b=-6$ together, we get $4b=-4 \implies b=-1$.solakis said:wait a minite if you substitute the 1st equation into the3rd don't you get b=2

- #8

solakis1

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My GOD i did this stupid substitution 3 timesKlaas van Aarsen said:If we add $a+b=2$ and $-a+3b=-6$ together, we get $4b=-4 \implies b=-1$.

An equation is a mathematical statement that shows the relationship between two or more quantities. It contains an equal sign (=) and typically involves variables, which represent unknown values, and constants, which are known values.

To solve an equation, you need to isolate the variable on one side of the equal sign and simplify the other side. This is done by using algebraic operations such as addition, subtraction, multiplication, and division. The goal is to get the variable by itself, with a coefficient of 1, on one side of the equal sign.

A polynomial equation is an equation that contains one or more terms with variables raised to non-negative integer exponents. The highest exponent in a polynomial equation is known as its degree. In the given equation, x^4+2x^3-x^2-6x-3=0, the highest degree is 4, making it a quartic polynomial equation.

A quartic equation can have up to four solutions, but it is not guaranteed. In some cases, there may be fewer or no real solutions. In the given equation, x^4+2x^3-x^2-6x-3=0, there may be up to four real solutions for x.

There are several methods for solving a quartic equation, including factoring, completing the square, using the quadratic formula, and using numerical methods such as graphing or iteration. In some cases, the equation may be simplified by using algebraic techniques before applying these methods.

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