Question

Find a cubic equation whose roots are the squares of those of x^3+px+q=0. State reasons and working

Answer 1

Do you know how to find a polynomial whose roots are negative of the given polynomial ?

Answer 2

not sure what you mean @ganeshie8

Answer 3

If r is a root if the given polynomial,
then you should find a polynomial that has a root as -r

Answer 4

so say thenormal poly was p(x) = (x-a)S(x)+R
the new poly would be p(x) = (x+a)S(x)+r

Answer 5

Assume the solutions are a,b, and c, So that \[(x-a)(x-b)(x-c)=0\]

The equation you're looking for is \[(x-a^2)(x-b^2)(x-c^2)=0\]

The solution of the original depressed cubic is known so you can look up the relation between
a,b,c, and p,q.
See

https://en.wikipedia.org/wiki/Cubic_function

Once you have the three solutions, plug them into the 2nd equation and multiply it out.

Answer 6

Nope. If r is a root of p(x), the you can factor p(x) as :

p(x) = (x-r)S(x)

Answer 7

yo @fwizbang i dont realy get wat ur tryna point out. could you explain differently> thanks

Answer 8

Yes. So let's first focus on funding a polynomial whose roots are negative of the given polynomial

Answer 9

how would we do that?

Answer 10

Every cubic polynomial has three roots, call them a,b, and c. Since the cubic is zero at each of the roots, we can write
\[x^3-px+q = (x-a)(x-b)(x-c)=0\]

If you write out the right hand side, you get \[a+b+c =0\]\[p=ab+bc+ac\] and

\[q=-abc\]

What happens to p and q if you change a,b, and c to -a,-b, and -c?

Answer 11

If r is a root of p(x),
then what's the value of p(x) at x= r ?

Answer 12

p is same so p is still ab+bc+ac and q becomes positive?

Answer 13

After solving I got this equation..

x3+ 2p(x2) +p2x+q2=0
Is it correct?

Answer 14

@Pawanyadav your answer is correct except it is -q^2 not +q^2 how did you do it?

Answer 15

Oh sorry writing mistake

Answer 16

how'd u get it

Answer 17

(x-a2)(x-b2)(x-c2)=0
fwizbang posted this above

Answer 18

then?

Answer 19

Do you have the roots(a,b,c). If you do, just multiply it out....

Answer 20

Just simplify it
And u will get..
x3-x2(a2+b2+c2)+x(a2b2+b2c2+c2a2)-a2b2c2=0

Answer 21

where did you get the x(a2b2... term

Answer 22

when i expanded i got:
x3-x2(a2+B2+C2)+x(a2b2)
did i do sth wrong

Answer 23

And from fwizbang equation
a+b+c=0
ab+bc+ca=p
abc=-q

So a2+b2+c2=-2p
And (a2b2+........+c2a2)=0
Just substitute this in above equation.

Answer 24

oh! i getit

Answer 25

Try again....

Answer 26

thanks!

Answer 27

Thanks u got it