Solving Quadratic Equations: Find p=-6

[Madhav Goel]

Homework Statement

3x^2 + px + 3 = 0, p>0, one root is square of the other, then p=?? [/B]

Homework Equations

sum of roots = -(coefficient of x)/(coefficient of x^2)[/B]
product of roots= constant term/coeffecient of x^2

The Attempt at a Solution

ROot 1 = a
root 2 = b
b=a^2
a.b= 3/3
a^3=1
a=1
a^2=b=1
a+b=-p/3=2
p=-6

Where did i go wrong??[/B]

 

[Ray Vickson]

Homework Statement

3x^2 + px + 3 = 0, p>0, one root is square of the other, then p=?? [/B]

Homework Equations

sum of roots = -(coefficient of x)/(coefficient of x^2)[/B]
product of roots= constant term/coeffecient of x^2

The Attempt at a Solution

ROot 1 = a
root 2 = b
b=a^2
a.b= 3/3
a^3=1
a=1
a^2=b=1
a+b=-p/3=2
p=-6

Where did i go wrong??[/B]

Nowhere: a = 1, p = -6 is the answer; it is the only real solution, but there are two other complex ones.

However, I think you need more explanation. Why do you set a^3 = 1? Why can you set a+b=2? These are true, but you need to give reasons for saying them. I think a better presentation would be to say $3 x^2 + px + 3 = 3(x-a)(x-a^2)$ and just expand out the factorization on the right.

 

[Mark44]

Nowhere: a = 1, p = -6 is the answer; it is the only real solution, but there are two other complex ones.

I also got a = 1, p = -6, but there is a condition given that p > 0.

However, I think you need more explanation. Why do you set a^3 = 1? Why can you set a+b=2? These are true, but you need to give reasons for saying them. I think a better presentation would be to say $3 x^2 + px + 3 = 3(x-a)(x-a^2)$ and just expand out the factorization on the right.

 

[Ray Vickson]

I also got a = 1, p = -6, but there is a condition given that p > 0.

OK: I missed that. If we allow complex roots (but insist on a real p) there is a value of p > 0 that works.

When I said p = -6 is the only real solution, that was misleading: it is the only solution in which both p and and the roots are real. There are others with real p but complex roots.

 

[Madhav Goel]

Nowhere: a = 1, p = -6 is the answer; it is the only real solution, but there are two other complex ones.

However, I think you need more explanation. Why do you set a^3 = 1? Why can you set a+b=2? These are true, but you need to give reasons for saying them. I think a better presentation would be to say $3 x^2 + px + 3 = 3(x-a)(x-a^2)$ and just expand out the factorization on the right.

OK: I missed that. If we allow complex roots (but insist on a real p) there is a value of p > 0 that works.

When I said p = -6 is the only real solution, that was misleading: it is the only solution in which both p and and the roots are real. There are others with real p but complex roots.

Someone said that I could use cube roots of unity?
What exactly is that??

 

[Ray Vickson]

Someone said that I could use cube roots of unity?
What exactly is that??

Google is your friend. Searching under the keyword 'cube roots of unity' turns up numerous hits; see, eg,
http://www.trans4mind.com/personal_development/mathematics/polynomials/roots.htm

 

[lurflurf]

There are two candidates for p one positive one negative, you found the wrong one.
3x^2 + px + 3 = 0, p>0
so the two roots are
$$w=\dfrac{-p+\sqrt{p^2-36}}{6} \text{ and } w^2=\dfrac{-p-\sqrt{p^2-36}}{6}$$
that is not the important part
we know as you argued above
w^3=1
p=-3(w+w^2)
can you use that to write p^2 in terms of p?
After that solve for p.