# Quartic equation

1. Mar 11, 2005

### p53ud0 dr34m5

this was on the aime, and i was wondering how to factor it without a calculator.

$$x^4-4x^3+6x^2-4x=2005$$

2. Mar 11, 2005

### shmoe

Do the coefficients 1, -4, 6, -4, look familiar? (think pascals triangle)

3. Mar 11, 2005

### p53ud0 dr34m5

will this method yield both imaginary and real roots? im not following your hint.

4. Mar 11, 2005

### shmoe

It will give all roots. Furthur hint:

$$x^4-4x^3+6x^2-4x+1=(x-1)^4$$

5. Mar 11, 2005

### p53ud0 dr34m5

ok, so:
i can add one to each side and have:
$$x^4-4x^3+6x^2-4x+1=2006$$
now, i can use the $(x-1)^4$
$$(x-1)^4=2006$$?

6. Mar 12, 2005

### saltydog

Yes. Next, use the formula for complex roots. You know:

$$\sqrt[4]{1}=+1,-1,+i,-i$$

So:

$$\sqrt[4]{2006}$$

is?

7. Mar 12, 2005

### p53ud0 dr34m5

thanks for your help. i have some more questions about other concepts, but ill post a new thread. im drained from getting up early for the sat. good thing it was easy.

8. Mar 12, 2005

### p53ud0 dr34m5

oh and how do you find the 4th root of a number without your calculator?

9. Mar 12, 2005

### Data

Well, you can get as close as you like to the positive real root by using Newton's Method:

We want the positive, real root of $$f(x)=x^4 - 2006$$. Since $$6^4 = 1296 < 2006 < 2401 = 7^4$$ we make a first guess that the root is $$x_1 = \frac{13}{2} = 6.5$$. Newton's method tells us that a better guess is

$$x_2 = x_1 - {f(x_1) \over f^\prime (x_1)} = \frac{13}{2} - \frac{\left(\frac{13}{2}\right)^4- 2006}{4\cdot \left(\frac{13}{2}\right)^3} =\frac{13}{2} - \frac{\frac{28561}{16} - \frac{32096}{16}}{\frac{8788}{8}} = \frac{13}{2} + \frac{3535}{17576} = \frac{114244}{17576} + \frac{3535}{17576} = \frac{117779}{17576} \approx 6.701$$

By iterating the method (using 6.701 as a second guess, and applying the same process) you can get arbitrarily close, since our function is continuous and the method does converge for this function. The root is really 6.6924129...

Last edited: Mar 12, 2005
10. Mar 12, 2005

### tongos

Let P be the product of the nonreal roots of x^4-4x^3+6x^2-4x=2005. Find the greatest integer that is less than or equal to P.

(1+fourthroot of 2006)(1-fourthroot of 2006)P= -2005
when you factor, you'll see that all roots have to multiply to give -2005

((2006^0.5)-1)P=2005

rationalizing the denominator, 2005(sqrt(2006)+1)/2005
sqrt(2006) is between 44 and 45. thus adding one will yield 45.(something). the greatest integer less than this is 45.

11. Mar 12, 2005

### tongos

i personally like the solution to number 8,

2^(333x-2)+2^(111x+2)=2^(222x)+1
has three real roots
find the sum of the roots.

let 2^(111x)=r

then r^3+16r-8r-4=0
let a,b,c be the roots of the polynomial.

then 2^(111x)=a x= Ina/(111In2)

then 2^(111x)=b x=Inb/(111In2)

then 2^(111x)=c x=Inc/(111In2)

by the third degree polynomial we know that the roots multiply to give 4.

thus (2In2)/(111In2)= 2/111 which is the sum of the roots. m/n, m+n=113

12. Mar 12, 2005

### p53ud0 dr34m5

see, i could get the answer to the one about the imaginaries with a calculator, but i couldnt without one. i suck without a calculator. the other problem you showed was really easy. i got that one right for sure, because i got 113, but i went about a different way to get the 2/111. i feel really really really stupid that i didnt think of newtons method to find the root. *sigh*

13. Mar 12, 2005

### tongos

there was no need for newton's method, thats why they say the greatest integer that is less than P. Though it did require to know that the sqrt(2006) is between 44 and 45.

Last edited: Mar 12, 2005
14. Mar 12, 2005

### tongos

i was surprised of the easiness of 1-12. 13,14, and 15 are killer problems though.

15. Mar 12, 2005

### p53ud0 dr34m5

heh, i was off by 3 on 15. it sucked. im not that good at math that requires so much thinking, because this is my first year actually doing any math competition. i was surprised that i even made above a 100 on the amc 12, haha. i think on 13 you have to use factorials, and i have no idea about 14. unfortunately, i thought a lot of th eproblems were hard. i only got 3 on the AIME right. i could have gotten at least a 6, but i can not add without a calculator. :rofl: and i was off by no more than 5 on three that i missed. oh, well. tongos, if i have a question about a problem, could i ask you. do you have aim?