Question

Can someone help me prove this
( I would type the ques ASAP)
@satellite73

Answer 1

If a and b be two real roots of the equation x^3 +px^2 + qx +r = 0 satisfying the relation ab+1=0 , then prove that r^2 +pr + q +1 = 0

Answer 2

@satellite73

Answer 3

So, considering c as the third root
\[\huge a + b + c = -p \]
\[\huge ab +ac + bc = q \]
\[\huge abc = -r \]

Answer 4

i got no idea, but doesn't the last equation tell you \(c=r\) ?

Answer 5

yes it does i am working , i will notify you when i get the answer

Answer 6

maybe that is all you need

Answer 7

factor the \(r\) out of
\[r^3+pr^2+qr+r=0\]

Answer 8

you get

\[r^2+pr+q+1=0\]

Answer 9

Wait i think i got it

Answer 10

you do

Answer 11

once you have \(r\) is a root, then it all works out nicely

Answer 12

yup!

Answer 13

We get c = r
and c(b+a) =q+1


So,

we have to prove
r^2 +pr + q +1 =0

c^2 -(a+b+c)c +bc+ac
c^2-ac -bc -c^2 +bc +Ac

Answer 14

i think you are doing too much work

Answer 15

you already said the third real root is \(r\) right? because
\[abc=-r\] and \(ab=-1\) making \(c=r\) so \(r\) is a root

Answer 16

that means \[r^3+pr^2+qr+r=0\]

Answer 17

which also tells you
\[r(r^2+pr+q+1)=0\] and since \(r\neq 0\) that makes
\[r^2+pr+q+1=0\]

Answer 18

nice!

Answer 19

Actually i have got little confidence doing these proving sums

Answer 20

you did all the work
it came down to figuring out that the third root was \(r\)

Answer 21

yes that's why i am so excited