# Thread: Solving an equation by substitution

1. ## Solving an equation by substitution

Hello, I've come across a problem in my textbook that really baffles me.

Task: to use substitution $\displaystyle u=x+1/x$ to solve the equation for x:

$\displaystyle x^2+3x-2+3/x+1/x^2=0$

It was easy to make the quadratic equation,

$\displaystyle u^2+3u-2=0$

$\displaystyle u1= (-3+sqrt(17))/2$
$\displaystyle u2=(-3-sqrt(17))/2$

but solving for x was a bit "ugly" (quoted because math is NOT ugly )

$\displaystyle x+1/x=(-3+sqrt(17))/2$
$\displaystyle x+1/x=(-3-sqrt(17))/2$

I've spent quite a time over this, any pointers?

2. Originally Posted by VImtermute
Hello, I've come across a problem in my textbook that really baffles me.

Task: to use substitution $\displaystyle u=x+1/x$ to solve the equation for x:

$\displaystyle x^2+3x-2+3/x+1/x^2=0$

It was easy to make the quadratic equation,

$\displaystyle u^2+3u-2=0$

$\displaystyle u1= (-3+sqrt(17))/2$
$\displaystyle u2=(-3-sqrt(17))/2$

but solving for x was a bit "ugly" (quoted because math is NOT ugly )

$\displaystyle x+1/x=(-3+sqrt(17))/2$
$\displaystyle x+1/x=(-3-sqrt(17))/2$

I've spent quite a time over this, any pointers?
Multiply through by $\displaystyle x$ and rearrange, to get quadratics.

RonL

3. Thanks for the reply CaptainBlack.
That's exactly what I did, but I ended up with a complex solution (i.e. "wrong") for $\displaystyle u1$ and square roots of square roots of whatever (rightfully wrong) for $\displaystyle u2$. Could you be a bit more verbose?

P.S.
The solution is supposed to be $\displaystyle -2-sqrt(3),-2+sqrt(3)$

4. Originally Posted by VImtermute
That's exactly what I did, but I ended up with a complex solution (i.e. "wrong") for $\displaystyle u1$ and square roots of square roots of whatever (rightfully wrong) for $\displaystyle u2$. Could you be a bit more verbose?

P.S.
The solution is supposed to be $\displaystyle -2-sqrt(3),-2+sqrt(3)$
I don't like the look of your roots to that quadratic in $\displaystyle u$.

I think that:

$\displaystyle x^2+3x-2+3/x+1/x^2=0$

gives:

$\displaystyle u^2+3u-4=0$

under the proposed substitution, which has roots $\displaystyle (u=-4) \vee (u=1)$

RonL

5. !!!

CaptainBlack, you're absolutely right. I always forget the rules for squaring polynomials

Thank you very much.

6. Originally Posted by VImtermute
That's exactly what I did, but I ended up with a complex solution (i.e. "wrong") for $\displaystyle u1$ and square roots of square roots of whatever (rightfully wrong) for $\displaystyle u2$. Could you be a bit more verbose?

P.S.
The solution is supposed to be $\displaystyle -2-sqrt(3),-2+sqrt(3)$
First off: There are four solutions to this equation. Two complex and two real. You have listed only the two real solutions.

$\displaystyle x^2+3x-2+3/x+1/x^2=0$

$\displaystyle u = x + \frac{1}{x}$
so
$\displaystyle u^2 = x^2 + 2 \frac{1}{x^2}$

So we wish to have
$\displaystyle au^2 + bu + c = ax^2 + bx + c + \frac{b}{x} + \frac{b^2}{x^2} = x^2 + 3x - 2 + \frac{3}{x} + \frac{1}{x^2}$

So
a = 1
b = 3
c = -4

Thus
$\displaystyle u^2 + 3u - 4 = 0$

$\displaystyle (u + 4)(u - 1) = 0$

So u = -4 or u = 1.

Thus
$\displaystyle x + \frac{1}{x} = -4$

$\displaystyle x^2 + 4x + 1 = 0$ ==> $\displaystyle x = -2 \pm \sqrt{3}$

and
$\displaystyle x + \frac{1}{x} = 1$

$\displaystyle x^2 - x + 1 = 0$ ==> $\displaystyle x = \frac{1 \pm \sqrt{3}}{2}$

-Dan
(Rats! FOILed again. )