Question

If \(\alpha\) & \(\beta\) are the roots of the equation \(3x^2-4x+1=0\) the equation whose roots are \(\dfrac{\alpha^2}{\beta}\),\(\dfrac{\beta^2}{\alpha}\) is:-

Answer 1

@sauravshakya @ParthKohli @UnkleRhaukus plz help :)

Answer 2

One way to do it is by finding alpha and Beta

Answer 3

\[\alpha\beta = \dfrac{1}{3}\]\[\alpha + \beta = \dfrac{4}{3}\]

Answer 4

And find:
(x-alpha^2/Beta) (x-Beta^2/Alpha) =0

Answer 5

Or solve the equation, then use the fact that \((x - s)(x - r)\) is the equation is \(s\) and \(r\) are the roots.

Answer 6

@jiteshmeghwal9 Didn't you read the `\dfrac` tip? =/

Answer 7

I did it as follows :-

\[3x^2-3x-x+1=0\]\[3x(x-1)-1(x-1)=0\]\[(3x-1)(x-1)=0\]\[\alpha=1, \beta=\frac{1}{3}\]\[\frac{\alpha^2}{\beta}=3\]\[{\beta^2 \over \alpha}=\frac{1}{9}\]\[3x^2+(3+\frac{1}{9})x+3.\frac{1}{9}\]

Answer 8

yaa..i read it @ParthKohli

Answer 9

It should be \(-\left(3 + \dfrac{1}{9}\right)\)

Answer 10

Everything else is fine =)

Answer 11

How did u gt a+b=4/3 ab=1/3 ?

Is there any other method to find the roots ?

Answer 12

if you observe the roots, the product of the new roots id \[(\alpha^{2}/\beta)*(\beta^{2}/\alpha)=\alpha \beta\] so the product of the roots doesnt change
now find the some of the roots
i.e. \[(\alpha^{3}+\beta^{3})/(\alpha \beta)=(\alpha+\beta)((\alpha+\beta)^{2}-3\alpha \beta)/(\alpha \beta)\]
with this you can find the sum of the roots
the quadratic equation is given by,
\[x^{2}-(\alpha ^{1}+\beta^{1})x+\alpha^{1}\beta^{1}\]
where alpha1 and beta1 are the new roots
this is a generalized solution.

Answer 13

I got that using Vieta's Formula. @jiteshmeghwal9 :P

Answer 14

\[(x-3)(x-\frac{1}{9})=x^2-x/9-3x+1/3\]

That's all i gt by putting the roots in the equation given by @sauravshakya but still i can't find the correct answer :(

Answer 15

That's it!

Answer 16

\[\Huge{\color{green}{\text{Ans.-}3x^2-32x+3=0}}\]

Answer 17

it's 3x^2-32x+3=0

Answer 18

then the answer given must be wrong :)

Answer 19

Yes, it's wrong.

Answer 20

so what's the answer ?

Answer 21

Did u get
alpha =1
and Beta =1/3

Answer 22

yup

Answer 23

Look @ my first reply

Answer 24

\[\left(x - \dfrac{1}{9}\right)\left(x -3\right)\]

Answer 25

U mean the answer that i gt from saurav's equation is the answer

Answer 26

@ParthKohli

Answer 27

the process that i quoted earlier is also on of the standard processes. you can follow it.
more over it is quite obvious that the product of the roots in the new equation must be the same as the earlier equation at 1/3. but the answers has a product of roots 1.

Answer 28

Yes. You can get the answer from Vieta's Formula too.

Answer 29

how ?

Answer 30

The new roots are \(\dfrac{1}{9}\) and \(3\). So the prod. is \(\dfrac{1}{3}\) and the sum is \(\dfrac{28}{9}\). Now use Vieta's Formula!

Answer 31

Thaanx... i gt it :)

Answer 32

\[x^2 - \dfrac{28}{9}x + \dfrac{1}{3} =0\]You can multiply both sides by \(9\) to get a better equation.

Answer 33

x^2-28x+3=0

Answer 34

Actually \(9x^2\) :P

Answer 35

right, LOL ;)