Question

If m, p and mp are the three real roots from x³ + mx² + mpx + p = 0 what's the sum of the roots?

a)1
b)-1
c)2
d)3

I've tried to build a system of three equations and I got p=0 but the exercise says that none of the roots can be 0. What should I do? Pls help me.

Answer 1

tricky lol.. thinking..

Answer 2

maybe @ash2326 knows this one

Answer 3

factor out something maybe..
x³ + mx² + mpx + p = 0
x²(x + m) + p(x + m) = 0

Answer 4

i think i got this.. maybe lol
(x+m)(x^2+p) = 0

Answer 5

x = -m
x^2 = -p, so p must be negative

Answer 6

substitute x = -m into the equation...let's see what we get..

Answer 7

x³ + mx² + mpx + p = 0
(-m)^3+m*(-m)^2 + m*p*(-m) + p = 0
-m^2*p + p = 0
p(1-m^2)=0

Answer 8

Melinda what's the sum of the roots if the roots are x, y and x
\[x^3+ax^2+bx+c=0\]
x+y+z=?

Answer 9

x³ + mx² + mpx + p = 0
x²(x + m) + p(x + m) = 0 => this doesnt look right

Answer 10

m = - (m + mp + p) @ash2326

Answer 11

yes it is melinda, pull out x^2 from the first two and then pull out a p from the last two

Answer 12

m+p+m*p=0
p(1-m^2)=0
so either p = 0, or 1-m^2 = 0, but none of the solutions can be 0, so m^2 = 1 is one of the solutions, or, m can be either -1 or 1

Answer 13

x^2 = -p, x = -m, so: (-m)^2 = -p

Answer 14

m^2 = -p
in either case, if u have 1 or minus one, m^2 is still 1, so -p = 1, or p=-1

Answer 15

So p =-1, m = +/- 1

Answer 16

m+p+mp =
If m is -1:
-1-1+(-1)(-1) = -2 + 1 = -1

Answer 17

If m is 1:
1+(-1)+(1)(-1) = 1 - 2 = -1

Answer 18

The sum is -1, elegant, eh? :D

Answer 19

Yeah I'm also getting -1 , did you understand @MelindaR?

Answer 20

you guys are awesome but actually i dont get it

Answer 21

Okay Melinda I'll try to go step by step.

Answer 22

I'm here, if you have any doubt, just ask for me

Answer 23

Bah, I would factor x³ + mx² + mpx + p = 0 as x²(x + m) + p(mx + 1)=0

Answer 24

Ohh shoot u're right, hold on then

Answer 25

Melinda let me guide you

Answer 26

the options are tricky, bah. xD ok ash

Answer 27

We have the roots as m, p and mp
so
\[m+p+mp=-m\]
\[mp+m^2p+mp^2=mp\]
and
\[m\times p\times mp=-p\]
from third equation we get
\[ m^2p^2+p=0 => p(m^2p+1)=0\]
p can't be zero
so
\[m^2p=-1\]
now in second equation we have
\[mp+m^2p+mp^2=mp\]
so we get
\[m^2p+mp^2=0\]
we have \[m^2p=-1\]
so
we get
\[mp^2=-1\]
so
\[m^2p=-mp^2=> m^2p+mp^2=0\]
we get
\[ mp(p+m)=0\]
mp can't be zero so
\[ m=-p\]
now use
\[m+p+mp=-m\]
m=-p
sp
\[m-m-m^2=-m\]
we get
\[ m^2-m=0\]
\[m(m-1)=0\]
m can't be zero so m=1 and p=-1
now
\[mp=1\times -1\]
\[mp=-1\]
now sum of roots
m+p+mp=1-1-1=-1

Answer 28

ash howd u get the first three equations?

Answer 29

its the theory of equations bah, check out this http://openstudy.com/study#/updates/4f4503f9e4b065f388dd4c1f

Answer 30

\[x^3+mx^2+mpx+p=(x-m)(x-p)(x-m p)\]
\[..................=(x^2-px-mx+m p)(x-m p)\]
\[..................=(x^2+(-p-m)x+m p)(x-m p)\]
\[..................=x^3+(-p-m)x^2+m p x-m p x^2-(-p-m)x m p -m^2p^2\]
\[.................=x^3+(-p-m-m p)x^2+(m p +p^2+m)x -m^2p^2\]
\[m=-p-m-m p ; m p=m p+p^2+m ; p=-m^2p^2\]

Answer 31

If I didn't make a mistake...

Answer 32

Thanks @MelindaR

Answer 33

Ohhh, I see lol i didn't know that.. yay for another random fun fact lol

Answer 34

all right myin's attacking the question.. so ima go back to probability

Answer 35

Thanks a lot, Bah!

Answer 36

Melinda did you understand?

Answer 37

lol i didnt rly help haha, but u're welcome

Answer 38

I'm trying ash, wait a sec pls

Answer 39

I did all of that as well, ash... I'm trying to see where did i go wrong

Answer 40

Take your time:)

Answer 41

Yah... Ash I really like your answer. I can't top that icecream with whoop cream. Great Job! :)

Answer 42

But at least I showed how to derive those equations you got :)

Answer 43

So in a way I put sprinkles on it. :)

Answer 44

Thanks @myininaya :D for extolling me. Yeah, you're the best, you made the ice cream heavenly:D

Answer 45

@ash2326 I dont get this.
so we get
m2p+mp2=0
we have
m2p=−1
so we get
mp2=−1

if we have m²p = -1, then how come that m²p + mp² = 0 isn't - 1 + mp² = 0 and consequently mp² = 1?

Answer 46

sorry I made a typo , you're right
\[m^2p=-1 , mp^2=1\]
and
\[m^2p+mp^2=0\]

Answer 47

@ash2326 Also, where did you get m² ? I only get that -m=p
now use
m+p+mp=−m
m−m−m2=−m

Answer 48

We have m+p+mp=-m
and we got m=-p pr p=-m
so
\[ m+(-m)+m(-m)=-m\]
we get
\[m-m-m^2=-m\]
or
\[-m^2=-m=>-m^2+m=0=>m^2-m=0\]

Answer 49

Oh, I can understand everything now. Thanks a lot @ash2326 and @myininaya !

Answer 50

Welcome @MelindaR glad to help:D :D

Answer 51

:)