Help finding roots

**starswept** · July 8th 2007, 09:18 AM

Hi, I need to solve for x in the following two equations:

(x-1/x)^2 - 77/12 (x-1/x) + 10 = 0

[I expanded out to get (x^2 + 1/x^2 - 77x/12 + 77/12x + 8). Then I used the factor theorem to plug in possible x's, and found that (x-4) is a factor. However, I don't know where to go from there, or if what I've done so far is correct.]

and

(3x-5)(3x+1)^2(3x+7) + 68 = 0

Thank you for any help!

**Jhevon** · July 8th 2007, 09:25 AM

Originally Posted by starswept

Hi, I need to solve for x in the following two equations:

(x-1/x)^2 - 77/12 (x-1/x) + 10 = 0

[I expanded out to get (x^2 + 1/x^2 - 77x/12 + 77/12x + 8). Then I used the factor theorem to plug in possible x's, and found that (x-4) is a factor. However, I don't know where to go from there, or if what I've done so far is correct.]

and

(3x-5)(3x+1)^2(3x+7) + 68 = 0

Thank you for any help!

don't expand! it is quadratic in (x - 1)/x

replace (x - 1)/x with another variable, say y, and you will see that

y^2 - (77/12)y + 10 = 0

solve for the roots of that equation, when you get them, replace y with (x - 1)/x and solve for x

EDIT: Oh, it was x - 1/x not (x - 1)/x. that does not change the process though

EDIT 2: by the way, you expanded incorrectly

EDIT 3: I must be tired. the expansion was correct, so scratch out EDIT 2. I still think my original plan is best though. Yes you have to do the quadratic formula 2 or 3 times (2 if you use Plato's method), but to that's easier than working with the quartic equation you would get if you continued down the path you were going

**Jhevon** · July 8th 2007, 09:50 AM

Originally Posted by janvdl

Let's use Jhevon's method.

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$y = \frac{(77/12) \pm \sqrt{(-77/12)^2 - 4(1)(10)}}{2}$

$y = \frac{(77/12) \pm \sqrt{1,1736}}{2}$

$y = 3,795 \ or \ y = 2,622$

incomplete, remember we want the x-values. so assuming what you did is correct (i have no time to check it) you now replace each y with x - 1/x and solve for x to find the corresponding x-values

**janvdl** · July 8th 2007, 09:59 AM

Originally Posted by janvdl

Let's use Jhevon's method... and the mathematician's best friend, the quadratic formula

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$y = \frac{(77/12) \pm \sqrt{(-77/12)^2 - 4(1)(10)}}{2}$

$y = \frac{(77/12) \pm \sqrt{1,1736}}{2}$

$y = 3,795 \ or \ y = 2,622$

$x - \frac{1}{x} = 3,795$

$x^2 - 3,975x - 1 = 0$

Use the quadratic formula again.

Then you'll see that x = 3.302 or
x = -0.302

$x - \frac{1}{x} = 2,622$

$x^2 - 2,622x - 1 = 0$

Use the quadratic formula AGAIN.

Then you'll get x = 2.414
x = -0.414

**Plato** · July 8th 2007, 10:07 AM

$y = \left( {x - \frac{1}{x}} \right)\quad \Rightarrow \quad 12y^2 - 77y + 120 = 0\quad \Rightarrow \quad \left( {3y - 8} \right)\left( {4y - 15} \right) = 0$

Now you solve for x:
$\left( {x - \frac{1}{x}} \right) = \frac{8}{3}\quad \& \quad \left( {x - \frac{1}{x}} \right) = \frac{{15}}{4}$

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