poly roots Q (1 Viewer)

[5647382910]

solve the equation 6x^4 - 11x^3 - 26X^2 + 22x + 24 = 0 given that the product of two of the roots is equal to the product of the other two

This is fitzpatrick: Q17 p173

thanks in advance

 

[shaon0]

solve the equation 6x^4 - 11x^3 - 26X^2 + 22x + 24 = 0 given that the product of two of the roots is equal to the product of the other two

This is fitzpatrick: Q17 p173

thanks in advance

I'll start you off. If you haven't started already.

Let roots be; a,b,c,d such that a+b=c+d.
Sum of roots:
a+b+c+d= -b/a
2(a+b)= 11/6
Thus, a+b= 11/12

Sum of roots (twice):
ab+ac+ad+bd+bc+cd= c/a
a(b+c+d)+b(c+d)+cd= -26/6
a(b(a+b))+b(a+b)+cd= -26/6
11ab/12 + 11b/12 +cd=-26/6
cd= -26/6-11b/12-11ab/12

 

[5647382910]

I'll start you off. If you haven't started already.

Let roots be; a,b,c,d such that a+b=c+d.
Sum of roots:
a+b+c+d= -b/a
2(a+b)= 11/6
Thus, a+b= 11/12

Sum of roots (twice):
ab+ac+ad+bd+bc+cd= c/a
a(b+c+d)+b(c+d)+cd= -26/6
a(b(a+b))+b(a+b)+cd= -26/6
11ab/12 + 11b/12 +cd=-26/6
cd= -26/6-11b/12-11ab/12

....... the product of the roots are equal not the sum

 

[clintmyster]

heres a start

ab = cd

a + b + c +d = -b/a = 11/6

abc + abd + acd + bcd = -d/a = -22/6 = -11/3

subbing ab = cd,

abc + abd + a(ab) + b(ab) = -11/3

ab(a + b + c + d) = -11/3

ab (11/6) = -11/3

ab = -2

or

abcd = e/a = 24/6 = 4
ab(ab) = 4
(ab)² = 4
ab = +-2 (now thats just plain weird =S)


and you can do the rest cos im lazy

 

[5647382910]

heres a start

ab = cd

a + b + c +d = -b/a = 11/6

abc + abd + acd + bcd = -d/a = -22/6 = -11/3

subbing ab = cd,

abc + abd + a(ab) + b(ab) = -11/3

ab(a + b + c + d) = -11/3

ab (11/6) = -11/3

ab = -2

and you can do the rest cos im lazy

LOL i already got that,
this is wat ive done:
ab = cd
a + b + c +d = 11/6
abc + abd + acd + bcd = -11/3
abcd = 4
ab = cd = -2 ( It cant = +2 as this wont satisfy a + b + c +d = 11/6)
(a + b)(c + d) = - 1/3

I cannot go further,
solutions are 4/3, -3/2, 1+ sqrt3, 1- sqrt3

 

[gurmies]

6x^4 - 11x^3 - 26x^2 + 22x + 24 = 0

ab = cd

a + b + c + d = 11/6

ab + ac + ad + bc + bd + cd = -13/3

abc + abd + acd + bcd = -11/3

abcd = 4

2ab + ac + ad + bc + bd = -13/3

abc + abd + a^2b + b^2a = -11/3

ab(a + b + c + d) = -11/3

ab = -2, therefore cd = -2

ac + ad + bc + bd = -1/3

a(c + d) + b(c + d) = -1/3

(a + b)(c + d) = -1/3

(a + b)[11/6 - (a + b)] = -1/3

Let u = a + b

6u^2 - 11u - 2 = 0

(6u + 1)(u - 2) = 0

u = -1/6 or/ u = 2

a + b = -1/6 or/ a + b = 2

-2/b + b = -1/6

-12 + 6b^2 = - b

6b^2 + b - 12 = 0

(3b - 4)(2b + 3) = 0

b = 4/3 and a = -3/2

-2 + b^2 = 2b

b^2 - 2b - 2 = 0

c = 1 + srqt3

d = 1 - sqrt3

 

[shaon0]

....... the product of the roots are equal not the sum

lol. Sorry about that. I misinterpreted the question.

 

[shaon0]

6x^4 - 11x^3 - 26x^2 + 22x + 24 = 0

ab = cd

a + b + c + d = 11/6

ab + ac + ad + bc + bd + cd = -13/3

abc + abd + acd + bcd = -11/3

abcd = 4

2ab + ac + ad + bc + bd = -13/3

abc + abd + a^2b + b^2a = -11/3

ab(a + b + c + d) = -11/3

ab = -2, therefore cd = -2

ac + ad + bc + bd = -1/3

a(c + d) + b(c + d) = -1/3

(a + b)(c + d) = -1/3

(a + b)[11/6 - (a + b)] = -1/3

Let u = a + b

6u^2 - 11u - 2 = 0

(6u + 1)(u - 2) = 0

u = -1/6 or/ u = 2

a + b = -1/6 or/ a + b = 2

-2/b + b = -1/6

-12 + 6b^2 = - b

6b^2 + b - 12 = 0

(3b - 4)(2b + 3) = 0

b = 4/3 and a = -3/2

-2 + b^2 = 2b

b^2 - 2b - 2 = 0

c = 1 + srqt3

d = 1 - sqrt3

Nice work

 

[Timothy.Siu]

lol i've done this question a few times

 

[H4rdc0r3]

6x^4 - 11x^3 - 26x^2 + 22x + 24 = 0

ab = cd

a + b + c + d = 11/6

ab + ac + ad + bc + bd + cd = -13/3

abc + abd + acd + bcd = -11/3

abcd = 4

2ab + ac + ad + bc + bd = -13/3

abc + abd + a^2b + b^2a = -11/3

ab(a + b + c + d) = -11/3

ab = -2, therefore cd = -2

ac + ad + bc + bd = -1/3

a(c + d) + b(c + d) = -1/3

(a + b)(c + d) = -1/3

(a + b)[11/6 - (a + b)] = -1/3

Let u = a + b

6u^2 - 11u - 2 = 0

(6u + 1)(u - 2) = 0

u = -1/6 or/ u = 2

a + b = -1/6 or/ a + b = 2

-2/b + b = -1/6

-12 + 6b^2 = - b

6b^2 + b - 12 = 0

(3b - 4)(2b + 3) = 0

b = 4/3 and a = -3/2

-2 + b^2 = 2b

b^2 - 2b - 2 = 0

c = 1 + srqt3

d = 1 - sqrt3

wow, wtf, how did you know to change c+d into -(a+b) and make it a quadratic.

 

[5647382910]

6x^4 - 11x^3 - 26x^2 + 22x + 24 = 0

ab = cd

a + b + c + d = 11/6

ab + ac + ad + bc + bd + cd = -13/3

abc + abd + acd + bcd = -11/3

abcd = 4

2ab + ac + ad + bc + bd = -13/3

abc + abd + a^2b + b^2a = -11/3

ab(a + b + c + d) = -11/3

ab = -2, therefore cd = -2

ac + ad + bc + bd = -1/3

a(c + d) + b(c + d) = -1/3

(a + b)(c + d) = -1/3

(a + b)[11/6 - (a + b)] = -1/3

Let u = a + b

6u^2 - 11u - 2 = 0

(6u + 1)(u - 2) = 0

u = -1/6 or/ u = 2

a + b = -1/6 or/ a + b = 2

-2/b + b = -1/6

-12 + 6b^2 = - b

6b^2 + b - 12 = 0

(3b - 4)(2b + 3) = 0

b = 4/3 and a = -3/2

-2 + b^2 = 2b

b^2 - 2b - 2 = 0

c = 1 + srqt3

d = 1 - sqrt3

cool,
the part i lost myself was not using the u substitution; ending up with a quartic equation
thanks heaps!
btw its pretty retarded how u can get 4 roots from one (i.e b)

 

[gurmies]

wow, wtf, how did you know to change c+d into -(a+b) and make it a quadratic.

I noticed that a + b was common and was an equation reducible to a quadratic

 

[Trebla]

solve the equation 6x^4 - 11x^3 - 26X^2 + 22x + 24 = 0 given that the product of two of the roots is equal to the product of the other two

This is fitzpatrick: Q17 p173

thanks in advance

This question looks so familiar...

I answered this EXACT question in an earlier thread lol:
http://community.boredofstudies.org/13/mathematics-extension-1/190810/surprisingly-hard-poly-q.html