1. ## collector's choice

I collect old school books. They have the advantage that you don't need calculators or computers to solve the problems published in those books.

In an arithmetic and algebra exercise book from 1883 I found a way to solve this equation exactly:

$x^4+2x^3-13x^2+2x+1=0$

Give it a try.

EB

2. Originally Posted by earboth
I collect old school books. They have the advantage that you don't need calculators or computers to solve the problems published in those books.

In an arithmetic and algebra exercise book from 1883 I found a way to solve this equation exactly:

$x^4+2x^3-13x^2+2x+1=0$

Give it a try.

EB
Perhaps this isn't what you had in mind, but note that the leading coefficient and constant terms are 1. So if this factors it must factor as:
$(x^2 + ax + 1)(x^2 + bx + 1) = x^4 + (a+b)x^3 + (ab + 2)x^2 + (a+b)x + 1$

So
$a + b = 2$
and
$ab + 2 = -13$

So $b = 2 - a$

$a(2 - a) = -15$

$a^2 - 2a - 15 = 0$

$(a - 5)(a + 3) = 0$

So a = -3 or a = 5.

The a = 5, b = -3 solution works. (So does a = -3, b = 5)

So
$x^4+2x^3-13x^2+2x+1 = (x^2 + 5x + 1)(x^2 - 3x + 1) = 0$

which we can now solve by setting the (irreducible) quadratic factors to 0.

-Dan

3. Originally Posted by topsquark
Perhaps this isn't what you had in mind...

-Dan
Hello topsquark,

first: It wasn't me who "invented" this method.

second: The following method only works if:
- the coefficients are arranged(?) symmetrically in the equation and
- $x \neq 0$

Then you can divide the equation by x²:

$x^4+2x^3-13x+2x+1=0$ Divide by x²
$x^2+2x-13+\frac{2}{x}+\frac{1}{x^2}=0$ Re-arrange:
$x^2+\frac{1}{x^2}+2x+\frac{2}{x}-13=0$ Factor:
$x^2+\frac{1}{x^2}+2 \left( x+\frac{1}{x} \right)-13=0$

Now set $z = x+\frac{1}{x}$. Then

$z^2 = x^2+2+ \frac{1}{x^2}$. That means

$z^2 -2= x^2+ \frac{1}{x^2}$

And now you can use substitution. The equation becomes:

$z^2-2+2 \cdot z-13=0$ which will give z = -5 or z = 3.

Re-substituate to solve for x

The actual solution for x doesn't matter. I only wanted to show a very special method to solve those equations.

EB

4. Originally Posted by earboth
Hello topsquark,

first: It wasn't me who "invented" this method.

second: The following method only works if:
- the coefficients are arranged(?) symmetrically in the equation and
- $x \neq 0$

Then you can divide the equation by x²:

$x^4+2x^3-13x+2x+1=0$ Divide by x²
$x^2+2x-13+\frac{2}{x}+\frac{1}{x^2}=0$ Re-arrange:
$x^2+\frac{1}{x^2}+2x+\frac{2}{x}-13=0$ Factor:
$x^2+\frac{1}{x^2}+2 \left( x+\frac{1}{x} \right)-13=0$

Now set $z = x+\frac{1}{x}$. Then

$z^2 = x^2+2+ \frac{1}{x^2}$. That means

$z^2 -2= x^2+ \frac{1}{x^2}$

And now you can use substitution. The equation becomes:

$z^2-2+2 \cdot z-13=0$ which will give z = -5 or z = 3.

Re-substituate to solve for x

The actual solution for x doesn't matter. I only wanted to show a very special method to solve those equations.

EB
Ah yes! I've seen that kind of trick pulled occasionally. I've got an older advanced algebra (HS level) algebra book that works with that kind of stuff. I had totally forgotten about that method.

-Dan