Question

if α,β are the roots of equation x^2-2x+3=0 then the equation whose roots are P=α^3-3α^2+5α-2 and Q=β^3-β^2+β+5 is

Answer 1

this is long...

Answer 2

find sum of roots, alpha + beta
and product alpha * beta first

Answer 3

then since we have P and Q, we simplify for P+Q and PQ

Answer 4

sum is 2 and product is 3

Answer 5

yes,

Answer 6

if we solve like dat do u think we wud reach to any answer as such

Answer 7

P+Q = a^3+b^3 -3a^2 - b^2+5a+b -2+5
no i am not sure, just trying...

Answer 8

okk..

Answer 9

there is an adjustment involved too

Answer 10

which advancement are u talking about

Answer 11

P+Q = a^3+b^3 -3a^2 - b^2+5a+b -2+5
=(a^3+b^3) - (a^2+b^2) +(a+b) - ( 2a^2 -4b+6) + 9

given a+b and ab
can you find
a^3+b^3
a^2+b^2
?

Answer 12

yaa i can find them out

Answer 13

then you'll get the value of P+Q
similarly try PQ

Answer 14

sum =-1

Answer 15

ok,
for 2a^2 -4b+6

since 'a' is the root
a^2-2a+3 = 0
so,
2a^2 = 4a -6
substitute this there...

Answer 16

so you'll need the value of a- b too

Answer 17

i hope there's some easier method for this...

Answer 18

i can find out the value of a-b

Answer 19

but did you get what i've done ?

Answer 20

yaa

Answer 21

i knew we could find a^3 +b^2 , a^2+b^2
thats why i made their groups and written them separately...

Answer 22

yaaa

Answer 23

oh i made an error

Answer 24

P+Q = a^3+b^3 -3a^2 - b^2+5a+b -2+5
=(a^3+b^3) - (a^2+b^2) +(a+b) - ( 2a^2 -4b+6) + 9

5a = a+4a
not a+4b

so,

P+Q = a^3+b^3 -3a^2 - b^2+5a+b -2+5
=(a^3+b^3) - (a^2+b^2) +(a+b) - ( 2a^2 -4a+6) + 9

now 2a62 -4a + 6 =0
you know why ?

Answer 25

2a^2 -4a + 6 =0

Answer 26

now you will get P+Q = 3 :)

Answer 27

try it and see if you get
a^3+b^3 = -10
a^2+b^2 = -2

Answer 28

4PQ=(P+Q)^2-(P-Q)^2
find P-Q and it gives you the value of PQ.

Answer 29

nice way to find PQ,
directly finding PQ would be nightmare...

Answer 30

haha..yea! :)

Answer 31

do u wanna say that we have to find out p-q also

Answer 32

instead of finding PQ, finding P-Q is easier
P-Q = a^3-b^3 -3a^2 + b^2+5a-b -2-5

try to group these , like i did ?

Answer 33

one group is obviously (a^3-b^3)

Answer 34

yaa

Answer 35

what about -3a^2+5a+b^2+b

Answer 36

-3a^2 = -a^2 -2a^2
5a = a +4a

Answer 37

and from original equation we know the value of -2a^2 +4a ...

Answer 38

-2a^2 +4a = - (2a^2-4a)
...

Answer 39

we have to take value of a-b for that because i m getting eqn as
\[a^3-b^3-(2a^2-4a+6)-(a^2+b^2)+a-b+1\]

Answer 40

now we have to put that underrot of -8 as value to solve for a^3-b^3 or do u have some other method

Answer 41

P-Q = a^3-b^3 -3a^2 + b^2+5a-b -2-5
= (a^3-b^3) + (b^2-a^2) + (a-b) - (2a^2-4a ) -7
no, i don't have another method which will avoid sqrt -8
...

Answer 42

shud we put value of a-b then to get ans

Answer 43

try it out, i am not sure..

Answer 44

\[(\sqrt{-8})(-2)-(2)(\sqrt{-8})+\sqrt{-8}-1\]

Answer 45

it wud give out to be
\[(-3\sqrt{-8}-1)\]

Answer 46

is it ryt uptil here

Answer 47

actually i am getting
\((\sqrt{-8})(1)-(2)(\sqrt{-8})+\sqrt{-8}-1\)
and the sqrt -8 is getting cancelled, as i suspected!

so, P-Q = -1
now use 4PQ=(P+Q)^2-(P-Q)^2
to find PQ
and you will get the required equation

Answer 48

the \(\large a^3-b^3= (a-b)((a^2+b^2) + ab)\)
a^2+b^2 was -2
and
ab = 3 so, you'll get only sqrt -8
and not -2sqrt -8
check it out...

Answer 49

yaa i got dat