HSC Mathematics Marathon (1 Viewer)

Drongoski · Feb 22, 2011

jyu said:
You've got the right answer. Care to share your solution method?

By brute force.

hscishard · Feb 22, 2011

Drongoski said:
By brute force.

Did you end up killing the equation?

Brute force a type of trial and error? It's a bit of a computer term

Drongoski · Feb 22, 2011

hscishard said:
Did you end up killing the equation?

Brute force a type of trial and error? It's a bit of a computer term

Brute force is not a computer term. Quite commonly used in maths solutions. It is often used for workman-like approach. No elegance whatsoever. I cross-multiplied the 2 sides and painstakingly cancelled out all the common terms - but I was very careful, numbering each pair I cancelled out so I could easily recheck should I falter.

I wonder what an elegant solution would look like.

hscishard · Feb 22, 2011

Oh..thought it might be related to the brute force method in hacking

Riproot · Feb 22, 2011

Drongoski said:
Brute force is not a computer term. Quite commonly used in maths solutions. It is often used for workman-like approach. No elegance whatsoever. I cross-multiplied the 2 sides and painstakingly cancelled out all the common terms - but I was very careful, numbering each pair I cancelled out so I could easily recheck should I falter.

I wonder what an elegant solution would look like.

I always do shit the hard way.
You say "Pascal's triangle" I say "7 sets of brackets"

Drongoski · Feb 22, 2011

hscishard said:
Oh..thought it might be related to the brute force method in hacking

I think more generally "brute force" refers to any approach to problem-solving that is crude like using a sledgehammer, unsubtle, unelegant.

Trebla · Feb 22, 2011

A substitution approach may save some of the algebraic pain

$\dfrac{W+10X}{Y+10Z}=\dfrac{W+X}{Y+Z}\\\\\Rightarrow (W+X+9X)(Y+Z)=(Y+Z+9Z)(W+X)\\\\\Rightarrow(W+X)(Y+Z)+9X(Y+Z)=(Y+Z)(W+X)+9Z(W+X)\\\\\Rightarrow X(Y+Z)=Z(W+X)\\\\\Rightarrow XY = ZW$

then expand and simplify

Drongoski · Feb 23, 2011

Trebla

Beautiful! Should've thot of that.

Drongoski · Feb 23, 2011

Post a new question please.

Riproot · Feb 23, 2011

Drongoski said:
Post a new question please.

Polynomials:
Express $\frac{3x^2-6x+10}{(x-4)(x^2+1)}$ as a sum of partial fractions.

Drongoski · Feb 23, 2011

Riproot said:
Polynomials:
Express $\frac{3x^2-6x+10}{(x-4)(x^2+1)}$ as a sum of partial fractions.

Let:

$\frac {3x^2 -6x+10}{(x-4)(x^2+1)}= \frac {A}{x-4} + \frac {Bx + C}{x^2+1}$

$\therefore 3x^2-6x+10 = A(x^2+1) + (Bx+C)(x-4)$

$x=4 \implies 3\cdot 16 - 6 \cdot 4 + 10 = 17A + 0 \implies a = 2$

$x = 0 \implies A - 4C = 10 \implies C = -2$

$x = 1 \implies 2A - 3B - 3C = 3-6+10 \implies 3B = 3 \implies B = 1$

$\therefore \frac {3x^2-6x+10}{(x-4)(x^2+1)} = \frac{2}{x-4} + \frac {x-2}{x^2+1}$

UnrealAnchovies · Feb 24, 2011

Given that two roots of the equation $54x^4 + 45x^3 - 204x^2 - 125x + 150 = 0$ are equal in absolute value but different in sign, write down the relations between the roots and coefficients and hence solve the equation.

hscishard · Feb 24, 2011

UnrealAnchovies said:
Given that two roots of the equation $54x^4 + 45x^3 - 204x^2 - 125x + 150 = 0$ are equal in absolute value but different in sign, write down the relations between the roots and coefficients and hence solve the equation.

5/3, -5/3, -18/12 = -3/2 and 8/12 = 2/3

Method, sub c and -c or whatever pronumeral you give it.
Is this as hard as polynomials get? I haven't done it yet

UnrealAnchovies · Feb 24, 2011

hscishard said:
5/3, -5/3, -18/12 = -3/2 and 8/12 = 2/3

Method, sub c and -c or whatever pronumeral you give it.
Is this as hard as polynomials get? I haven't done it yet

Not sure, haven't done most of polynomials. Though I presume it shouldn't be too challenging.

Omnipotence · Feb 24, 2011

Drongoski said:
Let:

$\frac {3x^2 -6x+10}{(x-4)(x^2+1)}= \frac {A}{x-4} + \frac {Bx + C}{x^2+1}$

$\therefore 3x^2-6x+10 = A(x^2+1) + (Bx+C)(x-4)$

$x=4 \implies 3\cdot 16 - 6 \cdot 4 + 10 = 17A + 0 \implies a = 2$

$x = 0 \implies A - 4C = 10 \implies C = -2$

$x = 1 \implies 2A - 3B - 3C = 3-6+10 \implies 3B = 3 \implies B = 1$

$\therefore \frac {3x^2-6x+10}{(x-4)(x^2+1)} = \frac{2}{x-4} + \frac {x-2}{x^2+1}$

Then integrate the function.

hscishard · Feb 24, 2011

Omnipotence said:
Then integrate the function.

Nop. The question just asked to express it in that form

Omnipotence · Feb 24, 2011

hscishard said:
Nop. The question just asked to express it in that form

I'm just extending that previous question. =]

Drongoski · Feb 24, 2011

Omnipotence said:
Then integrate the function.

$\int = 2ln \vert x-4 \vert + \frac{1}{2}ln (x^2 + 1) - 2tan^{-1} x + C$

HyperComplexxx · Feb 24, 2011

If z is a complex number such that z = r(cosθ +isinθ), where r is real, show that arg(z+r) = (1/2)θ .

ohexploitable · Feb 24, 2011

HyperComplexxx said:
If z is a complex number such that z = r(cosθ +isinθ), where r is real, show that arg(z+r) = (1/2)θ .

Draw it, it forms a rhombus, z+r is a diagonal, diagonals of rhombuses bisect angles.

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