# Thread: (x+1/x)^2-(3/2)(x-1/x)=4. Should I really solve a quartic equation

1. ## (x+1/x)^2-(3/2)(x-1/x)=4. Should I really solve a quartic equation

$\left(x+\frac{1}{x}\right)^2-\frac{3}{2}\left(x-\frac{1}{x}\right)$

When expanding the square, and adding the fractions I get a quartic equation. Is there any way?

2. ## Re: (x+1/x)^2-(3/2)(x-1/x)=4. Should I really solve a quartic equation

It may be a quartic equation in $x$, but it appears at first glance (without trying it out) to be a quadratic equation in $x^2$.

3. ## Re: (x+1/x)^2-(3/2)(x-1/x)=4. Should I really solve a quartic equation

Or: let $y= x+ \frac{1}{x}$, solve the quadratic equation $y^2- (3/2)y= 4$ for y, then solve $x+ \frac{1}{x}= y$ which will be another quadratic equation, for x. (The roots are NOT easy integer or even rational numbers.)

4. ## Re: (x+1/x)^2-(3/2)(x-1/x)=4. Should I really solve a quartic equation

Hello, NameIsHidden!

$\left(x+\frac{1}{x}\right)^2-\frac{3}{2}\left(x-\frac{1}{x}\right)\;=\;4$

Watch what happens when we expand . . .

$\quad\left(x+\frac{1}{x}\right)^2 - \frac{3}{2}\left(x - \frac{1}{x}\right) \;=\;4$

$x^2 + 2 + \frac{1}{x^2} - \frac{3}{2}\left(x - \frac{1}{x}\right) \;=\;4$

$x^2 - 2 + \frac{1}{x^2} - \frac{3}{2}\left(x - \frac{1}{x}\right) \;=\;0$

$\left(x -\frac{1}{x}\right)^2 - \frac{3}{2}\left(x - \frac{1}{x}\right) \;=\;0$

Let $u \,=\,x+\frac{1}{x}\!:\;\;u^2 - \tfrac{3}{2}u \:=\:0 \quad\Rightarrow\quad u\left(u - \tfrac{3}{2}\right) \:=\:0$

Hence: . $u \,=\,0,\;u \,=\,\tfrac{3}{2}$

Back-substitute:

. . $x + \frac{1}{x} \:=\:0 \quad\Rightarrow\quad x^2-1\:=\:0 \quad\Rightarrow\quad \boxed{x \:=\:\pm1}$

. . $x + \frac{1}{x} \:=\:\frac{3}{2} \quad \Rightarrow\quad 2x^2 - 3x - 2 \:=\:0 \quad\Rightarrow\quad (x-2)(2x+1) \:=\:0$

. . . . . . $\boxed{x \;=\;2.\:\text{-}\tfrac{1}{2}}$

5. ## Re: (x+1/x)^2-(3/2)(x-1/x)=4. Should I really solve a quartic equation

Originally Posted by HallsofIvy
Or: let $y= x+ \frac{1}{x}$, solve the quadratic equation $y^2- (3/2)y= 4$ for y, then solve $x+ \frac{1}{x}= y$ which will be another quadratic equation, for x. (The roots are NOT easy integer or even rational numbers.)
Thank you everybody. HallsofIvy, you should have seen the negative sign as positive in -3/2(x-1/x)

6. ## Re: (x+1/x)^2-(3/2)(x-1/x)=4. Should I really solve a quartic equation

Originally Posted by Matt Westwood
It may be a quartic equation in $x$, but it appears at first glance (without trying it out) to be a quadratic equation in $x^2$.
Sorry, ignore me, I was reading the second term as squared, and it appears it's not.

7. ## Re: (x+1/x)^2-(3/2)(x-1/x)=4. Should I really solve a quartic equation

Originally Posted by NameIsHidden
(x+1/x)^2-(3/2)(x-1/x)=4.
Dollars to donuts you missed a couple pairs of brackets and that should be:
[(x+1) / x]^2 - (3/2)[(x-1) / x] = 4.

Then you'll get a "nice" quadratic: 9x^2 - 7x - 2 = 0
and you'll happily end up with x = 1 or x = -2/9

Yes? No?

8. ## Re: (x+1/x)^2-(3/2)(x-1/x)=4. Should I really solve a quartic equation

Originally Posted by Wilmer
Dollars to donuts you missed a couple pairs of brackets and that should be:
[(x+1) / x]^2 - (3/2)[(x-1) / x] = 4.

Then you'll get a "nice" quadratic: 9x^2 - 7x - 2 = 0
and you'll happily end up with x = 1 or x = -2/9

Yes? No?
No. You are wrong. I did not miss out any brackets. I gave all the brackets necessary. Soroban got it right.

9. ## Re: (x+1/x)^2-(3/2)(x-1/x)=4. Should I really solve a quartic equation

Originally Posted by HallsofIvy
Or: let $y= x+ \frac{1}{x}$, solve the quadratic equation $y^2- (3/2)y= 4$ for y, then solve $x+ \frac{1}{x}= y$ which will be another quadratic equation, for x. (The roots are NOT easy integer or even rational numbers.)
The equation has the term : -(3/2) (x-1/x) ,so if you put x +1/x=y ,can you have :x-1/x =y??

10. ## Re: (x+1/x)^2-(3/2)(x-1/x)=4. Should I really solve a quartic equation

Originally Posted by Soroban
Hello, NameIsHidden!

Watch what happens when we expand . . .

$\quad\left(x+\frac{1}{x}\right)^2 - \frac{3}{2}\left(x - \frac{1}{x}\right) \;=\;4$

$x^2 + 2 + \frac{1}{x^2} - \frac{3}{2}\left(x - \frac{1}{x}\right) \;=\;4$

$x^2 - 2 + \frac{1}{x^2} - \frac{3}{2}\left(x - \frac{1}{x}\right) \;=\;0$

$\left(x -\frac{1}{x}\right)^2 - \frac{3}{2}\left(x - \frac{1}{x}\right) \;=\;0$

Let $u \,=\,x+\frac{1}{x}\!:\;\;u^2 - \tfrac{3}{2}u \:=\:0 \quad\Rightarrow\quad u\left(u - \tfrac{3}{2}\right) \:=\:0$

Hence: . $u \,=\,0,\;u \,=\,\tfrac{3}{2}$

Back-substitute:

. . $x + \frac{1}{x} \:=\:0 \quad\Rightarrow\quad x^2-1\:=\:0 \quad\Rightarrow\quad \boxed{x \:=\:\pm1}$

. . $x + \frac{1}{x} \:=\:\frac{3}{2} \quad \Rightarrow\quad 2x^2 - 3x - 2 \:=\:0 \quad\Rightarrow\quad (x-2)(2x+1) \:=\:0$

. . . . . . $\boxed{x \;=\;2.\:\text{-}\tfrac{1}{2}}$

u=x-1/x. am I right?
I think you made a typo once and did ctrl+c, ctrl+v. Anyway the answers are correct. Hence it must be a typo

11. ## Re: (x+1/x)^2-(3/2)(x-1/x)=4. Should I really solve a quartic equation

Originally Posted by NameIsHidden
u=x-1/x. am I right?
I think you made a typo once and did ctrl+c, ctrl+v. Anyway the answers are correct. Hence it must be a typo
Yes it is, because x+1/x does not imply x^2-1