Quartic Polynomial (1 Viewer)

[Dr. Zoidberg]

Need help again guys

Solve the equation 6x⁴-11x³-26x²+22x+24=0 given that the product of two of the roots is equal to the product of the other two.

The answers are

It's from the green 3U fitzpatrick book if anyone was wondering. P173

Thanks!

 

[bored of sc]

I'll have a crack.

let roots be A, B, C, D.

let F = AB, CD

ABCD = F² = e/a = 24/6 = 4 ------- (1)

A + B + C + D = -b/a = --11/6 = 11/6 --------- (2)

AB + AC + AD + BC + BD + CD = 2F + AC + AD + BC + BD = c/a = -26/6 = -13/3 -------------- (3)

ABC + ABD + ACD + BCD = FC + FD + AF + BF = -d/a = -22/6 = -11/3 =-------------------- (4)

(1) F² = 4
F = + 2

OH!

(4)
F (A + B + C + D) = -11/3
substitute (2) into (4)

F (11/6) = -11/3
F = -2

-2 = AB
-2 = CD

(3)
-4 + AC + AD + BC + BD = -13/3
AC + AD + BC + BD = -1/3

Too hard.

AB = CD
A = CD/B
B = CD/A
C = AB/D
D = AB/C
A = F/B
B = F/A
C = F/D
D = F/C

 

[Dr. Zoidberg]

Hmm, nice try!

Forgot to write the answers lol. The answers are

It's from the green 3U fitzpatrick book if anyone was wondering. P173

 

[lyounamu]

Hmm, nice try!

Forgot to write the answers lol. The answers are

It's from the green 3U fitzpatrick book if anyone was wondering. P173

I used the Newton's method of approximation to approximately determine the root. I went as far as -0.7. I should have determined the root by then.

 

[Dr. Zoidberg]

Damm, haven't learnt Newton's method yet. Teacher said that we'll be learning it next year.

 

[lyounamu]

Damm, haven't learnt Newton's method yet. Teacher said that we'll be learning it next year.

Well, you don't have to use Newton's method. As a matter of fact, bored of sc was quite close. He got: product of roots = 2 or -2.

You just had to work from there. Then you will have different quadratic equations where you can use the quadratic formula to find the values of the roots.

 

[lyounamu]

not a bad idea... you'll be able to get a pretty good guess out of newtons method for roots of polynomials if theyre nice enough. could be a good alternative to trial and error

I used the trial and error method first to get the right approximation.