Harder 3u polynomials Q (1 Viewer)

[addikaye03]

The equation x^3+ax^2+b=0 has three distinct roots, such that one is the sum of the other two. Show that a^3-8b=0

 

[tommykins]

回复: Harder 3u polynomials Q

let roots be x,y,z
but x = y+z (in q)


Sum of roots = -a
z+x+y = -a
2x = -a
x= -a/2

Sub this into the equation.

PS. I'm pretty sure it should be x^3 + ax^2 - b = 0

 

[addikaye03]

yer, factorise out one of the roots in the sum of two at a time and sub in the product of the other 2. Took me a while to think of though lol.. damn my feeble 3u mind huh

 

[midifile]

Re: 回复: Harder 3u polynomials Q

PS. I'm pretty sure it should be x^3 + ax^2 - b = 0

or prove a^3 + 8b = 0

 

[tommykins]

回复: Re: 回复: Harder 3u polynomials Q

Yeah that. Lol.

 

[addikaye03]

no it is def. typed up correctly

 

[namburger]

definitely a typo in the question then

 

[tommykins]

回复: Re: Harder 3u polynomials Q

Typo on their behalf then? Because you can't randomly make it -b.

 

[addikaye03]

the answer gets it as a^3-8b=0. i type up my solution if u really think it shouldn't be -b.. it works

 

[tommykins]

回复: Re: Harder 3u polynomials Q

Yeah type it then.

 

[addikaye03]

x+y+z=-a....(1)
xy+xz+yz=0....(2)
xyz=b...(3)

Since x=y+z therefore...
2x=-a or x=-a/2
sub into (3) gives -a/2(yz)=b
yz=-2b/a

therefore x(y+z)+yz=0
x(x)+(-2b/a)=0
x^2-2b/a=0
x^2=2b/a sub in (-a/2) for x

a^3-8b=0

We wouldn't get a question like that at 3u level would we, or q7?

 

[lyounamu]

x+y+z=-a....(1)
xy+xz+yz=0....(2)
xyz=b...(3)

Since x=y+z therefore...
2x=-a or x=-a/2
sub into (3) gives -a/2(yz)=b
yz=-2b/a

therefore x(y+z)+yz=0
x(x)+(-2b/a)=0
x^2-2b/a=0
x^2=2b/a sub in (-a/2) for x

a^3-8b=0

Nah, you are wrong, it should be xyz = -b

 

[addikaye03]

Nah, you are wrong, it should be xyz = -b

aww.. damn.. thanx guys. Me and the book must be wrong

 

[axlenatore]

Nah, you are wrong, it should be xyz = -b

yer cause the roots three at a time are found by -d/a.