rooty problem

**Aradesh** · February 11th 2006, 03:52 AM

hello i have this:
Cuberoot[6*Sqrt[3]+10] - Cuberoot[6*Sqrt[3]-10] = x

i know x to be 2. but i want to be able to show this. i have tried cubing both sides but everything just ends up turning into things where i can't get rid of the roots and such. i'm sure there must be a method to simplify it to 2.
ps. sorry i don't know how to make the pretty mathematical symbols that other people seem to be able to.
thanks for any help.

**topsquark** · February 11th 2006, 07:46 AM

Originally Posted by Aradesh

hello i have this:
Cuberoot[6*Sqrt[3]+10] - Cuberoot[6*Sqrt[3]-10] = x

i know x to be 2. but i want to be able to show this. i have tried cubing both sides but everything just ends up turning into things where i can't get rid of the roots and such. i'm sure there must be a method to simplify it to 2.
ps. sorry i don't know how to make the pretty mathematical symbols that other people seem to be able to.
thanks for any help.

My basic attack is to get the equation that the expression represents. There may be a simpler way to do it, but this is the way I grok things...

$x=(6\sqrt{3}+10)^{1/3}-(6\sqrt{3}-10)^{1/3}$
Cubing both sides:
$x^3=(6\sqrt{3}+10)-3(6\sqrt{3}+10)^{2/3}(6\sqrt{3}-10)^{1/3}$
$+3(6\sqrt{3}+10)^{1/3}(6\sqrt{3}-10)^{2/3}-(6\sqrt{3}-10)$

Express the second term as:
$(6\sqrt{3}+10)^{1/3}(6\sqrt{3}+10)^{1/3}(6\sqrt{3}-10)^{1/3}$
Then multiply the last two factors in this term:
$(6\sqrt{3}+10)^{1/3}(108-100)^{1/3}=2(6\sqrt{3}+10)^{1/3}$

Doing this to the third term as well, and simplifying, we get:
$x^3=20-6(6\sqrt{3}+10)^{1/3}+6(6\sqrt{3}-10)^{1/3}$
Rearranging a bit we get:
$(x^3-20)/6=-(6\sqrt{3}+10)^{1/3}+(6\sqrt{3}-10)^{1/3}$

Now, at the beginning of this, we defined the LHS of this to be -x, so
$(x^3-20)/6=-x$
or,
$x^3+6x-20=0$

Graphing this, we can easily see there is only one real root. Then, using whatever method you choose, we can find that the real solution is x=2.

-Dan

**CaptainBlack** · February 11th 2006, 08:38 AM

Originally Posted by topsquark

My basic attack is to get the equation that the expression represents. There may be a simpler way to do it, but this is the way I grok things...

$x=(6\sqrt{3}+10)^{1/3}-(6\sqrt{3}-10)^{1/3}$
Cubing both sides:
$x^3=(6\sqrt{3}+10)-3(6\sqrt{3}+10)^{2/3}(6\sqrt{3}-10)^{1/3}$
$+3(6\sqrt{3}+10)^{1/3}(6\sqrt{3}-10)^{2/3}-(6\sqrt{3}-10)$

Express the second term as:
$(6\sqrt{3}+10)^{1/3}(6\sqrt{3}+10)^{1/3}(6\sqrt{3}-10)^{1/3}$
Then multiply the last two factors in this term:
$(6\sqrt{3}+10)^{1/3}(108-100)^{1/3}=2(6\sqrt{3}+10)^{1/3}$

Doing this to the third term as well, and simplifying, we get:
$x^3=20-6(6\sqrt{3}+10)^{1/3}+6(6\sqrt{3}-10)^{1/3}$
Rearranging a bit we get:
$(x^3-20)/6=-(6\sqrt{3}+10)^{1/3}+(6\sqrt{3}-10)^{1/3}$

Now, at the beginning of this, we defined the LHS of this to be -x, so
$(x^3-20)/6=-x$
or,
$x^3+6x-20=0$

Graphing this, we can easily see there is only one real root. Then, using whatever method you choose, we can find that the real solution is x=2.

-Dan

Suspecting that $x=2$ is a solution suggest that we try dividing

$x^3+6x-20=0$

by

$x-2$ .

And indeed this works:

$x^3+6x-20=(x-2)(x^2+2x+10)=0$ ,

which confirms that $x=2$ is a root, and the discriminant of the
quadratic factor is negative and so there are no other real roots.

RonL

**Aradesh** · February 11th 2006, 09:21 AM

actually $x^3 + 6x = 20$ was what i was trying to solve in the first place, and got that messy result. without knowing x=2 is a solution to this, and without divine insight, how can you solve it?
thanks

**ticbol** · February 11th 2006, 10:00 AM

Originally Posted by Aradesh

hello i have this:
Cuberoot[6*Sqrt[3]+10] - Cuberoot[6*Sqrt[3]-10] = x

i know x to be 2. but i want to be able to show this. i have tried cubing both sides but everything just ends up turning into things where i can't get rid of the roots and such. i'm sure there must be a method to simplify it to 2.
ps. sorry i don't know how to make the pretty mathematical symbols that other people seem to be able to.
thanks for any help.

I don't know, but since there is no variable in the cuberoots in the lefthand side, why not just use the calculator to get x?

Were you told/asked to show the solution by expanding the equation?
If not, then, by calculator,
cubrt(20.39230485...) -cubrt(0.392304845....) = x
2.732050808..... -0.732050808.... = x
2 = x

That can be done "straight" or "continuous" or no need to write down partial findings. Meaning, manipulate the calculator until you get the 2 at the end.

**CaptainBlack** · February 11th 2006, 10:20 AM

Originally Posted by Aradesh

actually $x^3 + 6x = 20$ was what i was trying to solve in the first place, and got that messy result. without knowing x=2 is a solution to this, and without divine insight, how can you solve it?
thanks

Sketch the curve, this will tell you about where to look for the root. If
its close to an integer go ahead to see if its a root.

RonL

**ticbol** · February 11th 2006, 10:38 AM

Originally Posted by Aradesh

actually $x^3 + 6x = 20$ was what i was trying to solve in the first place, and got that messy result. without knowing x=2 is a solution to this, and without divine insight, how can you solve it?
thanks

Umm, I did not see this before I posted my answer above.
Anyway, for cubic equations there is a Cubic Formula you can use. Cardano's version of it is easier to use.

On this particular cubic equation, the "just by looking" works.
x^3 +6x = 20
x(x^2 +6) = 20
If x=2,
2(4+6) =? 20
Yes, so x=2.

**Aradesh** · February 11th 2006, 01:02 PM

thanks everyone for all the insight into this problem.
so i take it there is no way to do some algebra to have 'x=2' drop out? is that the general consensus? but rather, seeing that a result is probably 2 then checking that is the case after assuming this?

**ticbol** · February 11th 2006, 01:16 PM

Originally Posted by Aradesh

thanks everyone for all the insight into this problem.
so i take it there is no way to do some algebra to have 'x=2' drop out? is that the general consensus? but rather, seeing that a result is probably 2 then checking that is the case after assuming this?

I don't exactly get what you mean here, but take a look at Descarte's Rule of Signs when applied to your equation.
x^3 +6x = 20
x^3 +6x -20 = 0
See that lefthand side. There is onbly one change of sign. That from +6x to -20. That means, per the Rule, there is only one real root. Then, since 2 is a root, then there is no more real root left. So, since in a cubic equation there are exactly 3 roots, then the other 2 remaining roots have to be complex or i-roots.
Meaning, try as you might, even with divine intervention, you will not get another real number value of x, other than x=2, that will make that cubic equation true.

**topsquark** · February 11th 2006, 01:30 PM

My favorite method of choice is to first try the rational root method (I have temporarily forgotten the official name.)

Anyway, we have the equation:
$x^3+6x-20=0$
Now, list all the integer factors of the coefficient of the leading term:
$\pm1$ .
List the factors for the coefficient of the last term (i.e. the constant):
$\pm1,\pm2,\pm4,\pm5,\pm10,\pm20$ .

Now make a list of all the possible rational roots for the equation by dividing all the possibilities in the second list by all the possibilities in the first list. We get:
$\pm1,\pm2,\pm4,\pm5,\pm10,\pm20$ .
If the equation $x^3+6x-20=0$ has a rational root, it is in this list somewhere. (And obviously not all equations have a rational root, so other methods of solution are handy to have around.)

This method works on all polynomials equations with integer coefficients.

Anyway, we now have 12 possible rational roots. By plugging them in one at a time, we find that x=2 is the only rational root. By graphing you can see there is only one real root, so your expression can only be 2.

-Dan

**Aradesh** · February 12th 2006, 07:52 AM

Basically what i was after is a way of taking say
$(10+6\sqrt{3})^{1/3}-2/(10+6\sqrt{3})^{1/3}$ and simplifying it
or $(10+6\sqrt{3})^{1/3}-(-10+6\sqrt{3})^{1/3}$ as it is the same thing. like say i didn't know it was actually 2 (like not having a calculator to check it) i know you can turn it into a cubic equation, but i was just wondering if there was a way to reduce it to 2. i'm sure it must be possible. every way i try either leads me to the cubic equation or takes me around in circles

is the problem actually that since it is a cube root, there are three solutions, one real and two complex, where those three solutions are all solutions to the cubic: $x^3+6x-20=0$ ?

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