Quintic Roots (2 Viewers)

[Lukybear]

Quartic Roots

Forth degree equations are almost unsolvable for me. Please help!

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

 

[shaon0]

Re: Quartic Roots

Forth degree equations are almost unsolvable for me. Please help!

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

Let roots be; a,b,c,d.
ab=cd=-2.
a+b= 2
c+d= -1/6
After trying factor theorem more than 10 times. I get x=-1.5 or -3/2 as a root.
Checking solutions on my calculator:
x=3.5 isn't a root when subbing in for a or b. Thus c or d=4/3 is another root
Now factoring the poly:
P(x)=(2x+3)(3x-4)(x^2-2x-2)
Thus, a,b=1+-sqrt(3)

Probably a more quicker way to solve this equation. Ask me if more clarification is needed

 

[untouchablecuz]

man that was a b*tch to type

solve the equations to find roots

edit: beaten

 

[gurmies]

Quartic - not quintic xD

 

[shaon0]

man that was a b*tch to type

solve the equations to find roots

edit: beaten

My working follows the same. After this i tried factor theorem using rational roots theorem to find my pairs. I then subbed in x=-1.5 into either of the summation pairs for a+b or c+d. Then just factor the poly with the solutions you get for c and d.

 

[adomad]

would using the remainder theorem to find a factor and then factorising it by inspection help you find the roots?

 

[shaon0]

would using the remainder theorem to find a factor and then factorising it by inspection help you find the roots?

Yeah maybe, but it takes a while to find x=-3/2 and x=4/3 as roots as there are many possible pairings. But, idk it could work if you have that much time.

 

[Trebla]

From untouchablez's end of his working, you can easily deduce a pair of quadratic equations (which are then easily solvable) from the sum and products of each pair of roots.

 

[shaon0]

From untouchablez's end of his working, you can easily deduce a pair of quadratic equations (which are then easily solvable) from the sum and products of each pair of roots.

Oh, lol.

 

[Lukybear]

From untouchablez's end of his working, you can easily deduce a pair of quadratic equations (which are then easily solvable) from the sum and products of each pair of roots.

How so?

 

[gurmies]

 

[untouchablecuz]

another way:

x+y=k (1)
xy=c ---> x=c/y (2)

sub (2) in (1)

(c/y)+y=k
c+y²=yk
y²-ky+c=0

 

[shaon0]

Yeah, i usually do this but untouchablecuz's method is also good