Question

2x³ - x² + kx + 4 = 0
The multiplication of two of the roots is 1. What's the value of k?

a) -4
b) 4
c) 0
d) -8

How can I start solving this?

Answer 1

We have
\[ 2x^3-x^2+kx+4=0\]
Let's divide the whole equation by 2
we get
\[ x^3-\frac{x^2}{2}+\frac{k}{2} \times x+2=0\]
We know that standard cubic equation
\[x^3+ax^2+bx+c\]
if the roots are x, y , z
then
\[a=-(x+y+z)\]
\[b=(xy+yz+zx)\]
\[c=-xyz\]
Let the roots of given equation be x, y and z
then for the given equation
\[-1/2=-(x+y+z)\]
\[k/2=(xy+yz+zx)\]
\[2=-xyz\]
It's given that product of two roots is 1, let xy=1
We have\[2=xyz=>z=2\]
We have
\[ -1/2= -(x+y+z)=> x+y= -3/2\]
We have
\[k/2=(xy+yz+zx)\]
now we have xy=1, x+y=-3/2 and z=2
\[ k/2= 1+z(x+y)\]
so
\[k/2= 1+ 2\times \frac{-3}{2}\]
\[k/2=1-3\]
\[k/2=-2=>k=-4\]

Answer 2

Melinda did you understand?

Answer 3

I understood until the standard cubic equation...

Answer 4

See that's the theory of equations
Consider a quadratic equation
ax^2+bx+c=0
divide it by a
x^2+ b/a x + c/a=0
Let its roots be x and y
then
x+y=-b/a
or -b/a= sum of the roots
xy= c/a
or
c/a= sum of the roots taken 2 at a time
Say if we have a, b and c as roots, sum of roots taken 2 at a time will be
ab+bc+ca
Did you understand till now?

Answer 5

Melinda???

Answer 6

Wait a sec pls

Answer 7

sure!!

Answer 8

I'm following.

Answer 9

Now if you have a cubic equation with roots x, y and z
say
x^3+ax^2+bx+c=0
then a=-(sum of the roots)= -(x+y+x)
b= sum of the roots taken 2 at a time= (xy+yz+zx)
c= -(sum of the roots taken 3 at a time)= -(xyz)

Answer 10

That's the first time I'm hearing that. I'm writing it down. Interesting. Thank you so much!

Answer 11

Melinda this is valid for any type of equation
x^4+ax^3+bx^2+cx+d=0
Let its roots be w, x, y and z
then
a= -(w+x+y+z) sum of the roots
b=(wx+wy+wz+xy+xz+yz) sum of the roots taken 2 at a time
c=-(wxy+wxz+xyz+ywz) sum of the roots taken 3 at a time
d=(wxyz) sum of the roots taken 4 at a time

Answer 12

Did you understand it?

Answer 13

Oh I see. Yep, I understand it. I have another exercise which I suppose it's the same process. Check it out:
If m, p and mp are the 3 roots from x³ + mx² + mpx + p = 0, the sum of the roots will be...
Its the same thing, right?

Answer 14

yeah
m+p+mp=-m

Answer 15

Thanks! I will try to solve it.

Answer 16

Welcome :)

Answer 17

I mean this:

2=−xyz
It's given that product of two roots is 1, let xy=1 We have
2=xyz=>z=2

Answer 18

hey @MelindaR
it's because
when
we have
\[x^3+ax^2+bx+c=0\]
let the roots be x, y and z
then
\[ x+y+z=-a\]
\[xy+yz+zx=b\]
\[xyz=-c\]
remember that signs change, beginning with -, then + and so
here c=4
Sorry I made a mistake
it'd be
xyz=-4
so
z=-2
sorry @MelindaR

Answer 19

That's ok. But I've tried with z=-2 and I didn't reach any results from the options...

Answer 20

Let me check once

Answer 21

I got k=-8, what did you get?

Answer 22

We have
\[2x^3-x^2+kx+4=0\]
Divide the whole equation by 2
\[x^3-\frac{1}{2}x^2+\frac{k}{2} x+2=0\]
let the roots be a, b and c.
\[ a+b+c=\frac{1}{2}\]
\[ab+bc+ca=\frac{k}{2}\]
and
\[abc=-2\]
Let ab =1
we get c=-2
so
\[a+b+c=\frac{1}{2}\]
c=-2
we get
\[a+b=\frac{5}{2}\]
now
\[ab+bc+ca=\frac{k}{2}\]
ab=1
\[1+c(b+a)=\frac{k}{2}\]
\[b+a=-\frac{5}{2}\]
and
c=-2
so

\[1-2\times \frac{-5}{2}=\frac{k}{2}\]
we
get
\[1-5=\frac{k}{2}\]
we get
\[k=-8\]

Answer 23

Mmm I made a silly mistake in the end. Thx a lot