Two roots of the equation 2x^3 - 3x^2 + px + q are 3 and -2. Find the third root of the rquation.
2006-09-17 00:01:07 · 4 answers · asked by Sasha 2 in Science & Mathematics ➔ Mathematics
x=1/2
From x=3 we get 54-27+3p+q=0
1) 3p+q=-27
From x=-2 we get -16-12-2p+q=0
2) -2p+q=28
Solving 1) and 2) simultaneously we get
p=-11 and q=6 so the original equation is
3) 2x^3-3x^2-11x+6=0 and since 2 roots are 3 and -2
we can factor 3) into (x-3)(x+2)(x+r)=
2x^3-3x^2-11x+6
Now with polynomial division divide both sides by
(x-3)(x+2) which is x^2-x-6 to solve for x-r and we get
2x-1. So 2x-1=0
x=1/2
Checking 2(1/2)^3-3(1/2)*2-11/2+6=
1/4-3/4-11/2+6=
-12/2+6=-6+6=0 OK
2006-09-17 00:33:38 · answer #1 · answered by albert 5 · 1⤊ 0⤋
2
2006-09-17 07:11:40 · answer #2 · answered by George 3 · 0⤊ 1⤋
how is it an equation? 2x^3-3x^2+px+q equal what?
2006-09-17 09:00:01 · answer #3 · answered by Carpe Diem (Seize The Day) 6 · 0⤊ 0⤋
if 3,-2,a are the roots, then
3-2+a = 3/2
hence a=1/2.... so straight....
2006-09-17 07:05:17 · answer #4 · answered by m s 3 · 0⤊ 1⤋