View Full Version : Can't isolate x!!...
05-10-02, 12:51 AM
I have this equation, but I can't find x! Can someone help pretty please? I'm going nuts trying to unravel this. :eek::bugeye:
I'm trying to learn the algebra required; I can easily find it with graphing software. I need to understand the algebra behind it...
I'll show you my attempt, but I have my doubts about it.
y = 0.25(x-2)^3 + (x-2)^2 - (x-2) + 2
I'm trying to find the x-intercept, so I let y equal zero, as usual...
0 = 0.25(x-2)^3 + (x-2)^2 - (x-2) + 2
and I start working on it...
0 = 0.25x^3 - 1.5x^2 + 3x - 2 + x^2 - 4x + 4 - x + 4
0 = 0.25x^3 + 0.5x^2 - 2x + 6
I get rid of fractions by multiplying all terms by 4
(4)0 = (4)0.25x^3 + (4)0.5x^2 - (4)2x + (4)6
0 = x^3 + 2x^2 - 8x + 24
I do a little factoring...
-24 =x(x^2 + 2x^1 - 8)
-24 =x(x - 4)(x + 2)
the closest equation I can get is
0 =x(x - 4)(x + 2) + 24
but my I don't know how to isolate x by itself on one side of the = sign... I don't know what to do with 24.... :(:mad:
I know there's only one x intercept, and it's very near to
x = - 3.0925, but that's by reading manually off the graph attached below. ( I love GrafEq (www.peda.com) )
What algebra do I need to do get this number?
The equation is a cubic polynomial, you can't easily factor x by itself. What you need to do is solve for the roots of the polynomial and go from there. I am sure your text book will have the general technique for finding polynomial roots.
05-10-02, 01:10 AM
Ok im stupid and if you read my post ignore it
05-10-02, 01:29 AM
Well, I looked up "roots" in my Calculus book, and basically it told me how to find x by trial and error...
Essentially I'm supposed to just keep testing numbers, with a little educated guess of which number I'm hunting...... "In general, Newton's method is to construct a sequence of increasingly accurate approximations..." Sigh. I was hoping for some magic trick to isolate x. It's not fair of my teacher to have us find the x-intercept among twelve other difficult things in an hour test ... :(
05-10-02, 01:40 AM
Are you alowed to use graphics calculators?
that will do it very simply
05-10-02, 02:02 AM
Arghhh! This is a hard one to do analytically.
However, there is a general solution for cubic equations. I suggest you do a google search for "solution cubic equation"
Yep, there are numerous methods for finding polynomial roots. Newtons method is but one. Without knowing what postitlord has been taught, and what method the teacher is testing for, it's hard to say which is the correct method.
The general solution suggested by James R was the one I was thinking of.
05-12-02, 04:40 PM
So basically I'm being told when x gets a power bigger than 1, there's no way to get one term of x by itself :(:(
x = stuff / stuff^42 * 0.175 - e
I just made that up, even if the stuff on the right side of the equation is ugly.
Are you alowed to use graphics calculators? Not in a test, but outiside sure. But as this was part of a test question, I'm trying to learn any way better than substitute x thirty times to approximate the answer...
I'm looking up "solution cubic equation" now... Hope I get something SIMPLE to understand... But I don't understand what I'm finding :bugeye: :rolleyes:
05-12-02, 10:01 PM
I believe you can get an analytic solution in the form x=something for equations where the highest power is anything up to 4. Higher than that becomes difficult. So, linear, quadratic, cubic and (I think) quartic equations have analytic solutions. You probably need numerical methods for higher powers.
when u get x to a power bigger than 1, there becomes more than one answer for x. thus x= is not a possiblety. and x^3 means either no roots, the same roots or differnt roots. but the number on the x axis can be found using another equation over the top of ur graph to find the points. this is above me tho.
05-14-02, 12:34 AM
An analytical solution is impossible for polynomials of order 5 or more. The analytical solutions for 3rd & 4th order polynomials are so messy, that nobody uses them except for students taking certain math courses.
The Newton method can be used to find the roots of almost any analytical function. For polynomials, it is pretty easy to use if you have a decent calculator.
For the polynomial you have, make up a new polynomial by substituting v = (x-2). When you find a value of v, x = (v+2).
A 3rd order polynomial has at least one real root. If you find it, you can use (x-root)(quadratic) = cubic. Solve the quadratic to find the other two roots. Newton is messy if you must use complex arithmetic to find the complex roots.
The general formula is the following (just guess at a value for Thisx).
Nextx = Thisx - Polynomial(Thisx) / Derivative(Thisx)
After doing the above, use the value of Nextx for Thisx. Keep this up until Nextx is pretty close to Thisx.
The formula for the derivative of a polynomial can be worked out from the following example.
P(x) = ax^3 + bx^2 + cx + d
D(x) = 3ax^2 + 2bx + c
For each term, multiply the coefficient by the exponent and subtract one from the exponent.
Newton converges pretty fast.
For the special cubic x^3 - 8 = 0 (cube root of 8), you would get the following if you guessed three for Thisx. Once you get close, it zooms in.
3.000 000 000
2.296 296 296
2.036 587 403
2.000 653 359
2.000 000 213
2.000 000 000
I picked the above example because I have a program for my HP calculator which shows the successive approximations for finding nth roots using the Newton method.
I hope the above is understandable and that it helps. If not, post some questions, and I will try to answer them.
First, you have to get it into the formula:
0= ax<SUP>3</SUP>+bx<SUP>2</SUP>+cx+d (set y to 0)
-turn the x-2 into x
how do you do this?
(x-2)(x-2)=x<SUP>2</SUP>-4x+4 etc. I'll assume you know how to do that.
Now, after you do that, use this equation to solve:
f = [(3c/a) - (b²/a²)]/ 3
g = [(2b³/a³) - (9bc/a²) + (27d/a)]/27
h = (g²/4) + (f³/27)
If h > 0, there is only 1 real root and is solved by another method.
For the special case where h = 0, all 3 roots are real and equal.
When h <= 0, as is the case here, all 3 roots are real and we proceed as follows:
i = (g²/4) - h)<SUP> ½ </SUP>
j = (i)<SUP>1/3</SUP>
k = arc cosine (- (g / 2i))
L = j * -1
M = cosine (K/3)
N = (Square Root of 3) * sine (K/3)
P = (b/3a) * -1
x<SUB>1</SUB> = 2j * cosine(k/3) -(b/3a)
x<SUB>2</SUB> = L * (M + N) + P
x<SUB>3</SUB> = L * (M - N) + P
Now that you have a thorough reveiw of the alphabet, here is the method for when only 1 root is real:
U = (T)<SUP>1/3</SUP>
X<SUB>1</SUB> = (S + U) - (b/3a)
X<SUB>2</SUB> = -(S + U)/2 - (b/3a) + i*(S-U)*(3)<SUP>½</SUP>/2
X<SUB>3</SUB>= same thing - i*(S-U)*(3)½/2
When all three roots are real and equal:
x<SUB>1</SUB> = x<SUB>2</SUB> = x<SUB>3</SUB>= (d)<SUP>1/3</SUP> * -1
You're all making this FAR too complicated.
This is a cubic, but if you take out a factor of (x-2) you get a factor and hence a root straight away and the equation becomes that factor times some quadratic, and everyone can solve quadratics right?
So you get:
so you can see that this =0 when x=2 and/or when the quadratic =0.
Then solve the quadratic using the formula:
x-2 = -b +- sqrt(b^2 - 4ac)
this gives you:
x-2 = -1 +- 0
both answers are -2, so x-2 = -2 therefore x=0
So, x=2 and x=0
On Radioactive Waves
05-22-02, 06:02 PM
Newton converges pretty fast.
there are some cases when it dosn't......
05-22-02, 09:14 PM
You've made a mistake, and your answer is wrong. Plug x=0 into the original equation and you do not get y=0.
I know, I suck. I didn't take the factor out of 2.
I hang my head in shame and humbly apologise for any offense and/or inconvenience this may have caused to you or your family.
:( :( :( :( :(