HSC 2014 MX2 Marathon (archive) (3 Viewers)

glittergal96 · Oct 14, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $Given$ \ a,b,c \ $are real$ \\ $Show that at least one of the following equations has at least one real root$ \\ \\ x^2 + (a-b)x + (b-c) = 0 \\ x^2 + (b-c)x + (c-a) = 0 \\ x^2 + (c-a)x + (a-b) = 0$

We have

$p(0)+q(0)+r(0)=(b-c)+(c-a)+(a-b)=0.$

so at least one term on the LHS must be non-positive. As each of these polynomials is concave up, at least one must be concave up with vertex on below the x-axis. This polynomial must then have at least one real root.

glittergal96 · Oct 14, 2014

Re: HSC 2014 4U Marathon

glittergal96 said:
We have

$p(0)+q(0)+r(0)=(b-c)+(c-a)+(a-b)=0.$

so at least one term on the LHS must be non-positive. As each of these polynomials is concave up, at least one must be concave up with vertex on below the x-axis. This polynomial must then have at least one real root.

Here is a perhaps cleaner way of presenting this same argument:

Suppose each of these polynomials has no real roots. As they are concave up, this means that they are positive definite.

Then the sum of the three polynomials must also be positive definite.

But the sum of the three quadratics by direct calculation is $3x^2$ , which is NOT positive definite. Contradiction.

Tugga · Oct 14, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $Given$ \ a,b,c \ $are real$ \\ $Show that at least one of the following equations has at least one real root$ \\ \\ x^2 + (a-b)x + (b-c) = 0 \\ x^2 + (b-c)x + (c-a) = 0 \\ x^2 + (c-a)x + (a-b) = 0$

Could you also do this:

Assume all of the eqns have non-real complex roots. Then by conjugate root theorem and summing product of roots you get a sum of real squares = 0.

Therefore, all the roots must be zero, a contradiction of the assumption.

Axio · Oct 15, 2014

Re: HSC 2014 4U Marathon

Tugga said:
Could you also do this:

Assume all of the eqns have non-real complex roots. Then by conjugate root theorem and summing product of roots you get a sum of real squares = 0.

Therefore, all the roots must be zero, a contradiction of the assumption.

Do all the roots have to be zero just because the sum of their products is zero?
__________
And could somebody please post another complex number question, thanks.

Tugga · Oct 15, 2014

Re: HSC 2014 4U Marathon

Axio said:
Do all the roots have to be zero just because the sum of their products is zero?
__________
And could somebody please post another complex number question, thanks.

When you sum the product of roots you get this: (gonna try use fancy latex)

$$Let the non-real, complex roots be $ \alpha, \bar{\alpha}, \beta, \bar{\beta}, \gamma \bar{\gamma} \\ $Then by summing the product of roots from each equation$ \\ \alpha \bar{\alpha} + \beta \bar{\beta} + \gamma \bar{\gamma} = 0 \\ |\alpha|^2 + |\beta|^2 + |\gamma|^2 = 0 \\ $Since $|\alpha|, |\beta|, |\gamma|$ are all real$\\ $Therefore, $|\alpha|=|\beta|=|\gamma|=0$ \\ Contradiction$

Axio · Oct 15, 2014

Re: HSC 2014 4U Marathon

Tugga said:
When you sum the product of roots you get this: (gonna try use fancy latex)

$$Let the non-real, complex roots be $ \alpha, \bar{\alpha}, \beta, \bar{\beta}, \gamma \bar{\gamma} \\ $Then by summing the product of roots from each equation$ \\ \alpha \bar{\alpha} + \beta \bar{\beta} + \gamma \bar{\gamma} = 0 \\ |\alpha|^2 + |\beta|^2 + |\gamma|^2 = 0 \\ $Since $|\alpha|, |\beta|, |\gamma|$ are all real$\\ $Therefore, $|\alpha|=|\beta|=|\gamma|=0$ \\ Contradiction$

Ah k, yeah it looks good to me.

Axio · Oct 26, 2014

HSC 2015 4U Marathon

New Question (Cambridge):

If |z| = r and arg z = theta, show that z/[(z^2)+(r^2)] is real and give its value.

Chlee1998 · Oct 31, 2014

Re: HSC 2014 4U Marathon

You change r to mod z. Then factor z out of denominator which cancels with the top. Then u get 1/ 2rcos theta

EDIT: did u type the question wrongly or did Realisenothing read it wrong?

RealiseNothing · Oct 31, 2014

Re: HSC 2015 4U Marathon

Axio said:
New Question (Cambridge):

If |z| = r and arg z = theta, show that z/[(z^2)+(r^2)] is real and give its value.

$\frac{z^2+|z|^2}{z} = z+\frac{z\bar{z}}{z} = z+\bar{z}$

Chlee1998 · Oct 31, 2014

Re: HSC 2015 4U Marathon

RealiseNothing said:
$\frac{z^2+|z|^2}{z} = z+\frac{z\bar{z}}{z} = z+\bar{z}$

wasn't z on the numerator?

glittergal96 · Nov 1, 2014

Re: HSC 2015 4U Marathon

Chlee1998 said:
wasn't z on the numerator?

The inverse of a nonzero real number is real.

Chlee1998 · Nov 1, 2014

Re: HSC 2015 4U Marathon

glittergal96 said:
The inverse of a nonzero real number is real.

Yes, but what's the point of doing it Realise's way? like in the end he still has to find the real value. I don't think there's anything wrong with my solution? (in the post after the question)

glittergal96 · Nov 1, 2014

Re: HSC 2014 4U Marathon

No-one said there was anything wrong with your solution, he just wrote up what is logically the same thing in a different notation. (With the typo of having the reciprocal of what we actually want, and he didn't finish up by writing $2r\cos(\theta)$ .)

I just meant that its pretty obvious what he meant, so picking on the typo seemed a bit petty.

Sy123 · Nov 1, 2014

Re: HSC 2014 4U Marathon

$\\ $Find the exact form of$ \ \cos \frac{\pi}{10}$

integral95 · Nov 1, 2014

Re: HSC 2014 4U Marathon

Sy123 said:
$\\ $Find the exact form of$ \ \cos \frac{\pi}{10}$

Using a calculator we have

$cos\frac{\pi}{5} + cos\frac{3\pi}{5} = \frac{1}{2} \\ \\ cos\frac{\pi}{5}cos\frac{3\pi}{5} = -\frac{1}{4} \\ $We use those to make a quad equation containing those roots, then take the quadratic formula and deduce that$ cos\frac{\pi}{5}$ is the bigger value, then using the double angle formula you get your solution$$

Sy123 · Nov 1, 2014

Re: HSC 2014 4U Marathon

integral95 said:
calculator

ಠ_ಠ

RealiseNothing · Nov 2, 2014

Re: HSC 2014 4U Marathon

glittergal96 said:
No-one said there was anything wrong with your solution, he just wrote up what is logically the same thing in a different notation. (With the typo of having the reciprocal of what we actually want, and he didn't finish up by writing $2r\cos(\theta)$ .)

I just meant that its pretty obvious what he meant, so picking on the typo seemed a bit petty.

Wasn't a typo, I just flipped it since like you said the inverse of a real number is still real, and I thought it was easier to manipulate when it was flipped compared to the original question.

RealiseNothing · Nov 2, 2014

Re: HSC 2014 4U Marathon

RealiseNothing said:
Wasn't a typo, I just flipped it since like you said the inverse of a real number is still real, and I thought it was easier to manipulate when it was flipped compared to the original question.

Though looking back on it, there really isn't much difference in flipping it or not lol.

glittergal96 · Nov 2, 2014

Re: HSC 2014 4U Marathon

RealiseNothing said:
Though looking back on it, there really isn't much difference in flipping it or not lol.

nope, none lol.

Sy123 · Nov 2, 2014

Re: HSC 2014 4U Marathon

$\\ (0,0), (a,11) \ $and$ \ (b,37), \ $are vertices of an equilateral triangle, find the value of$ \ ab$

HSC 2014 MX2 Marathon (archive) (3 Viewers)

Active Member

Active Member

Member

=o

Member

=o

=o

Member

what is that?It is Cowpea

Member

Active Member

Member

Active Member

This too shall pass

Well-Known Member

This too shall pass

what is that?It is Cowpea

what is that?It is Cowpea

Active Member

This too shall pass

Users Who Are Viewing This Thread (Users: 0, Guests: 3)