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Re: HSC 2015 4U Marathon $The equation$ \ x^3+3x+2 \ $has roots \alpha , \beta,\gamma .$ $Find the equation with roots \alpha +\frac{1}{\alpha }, \beta +\frac{1}{\beta }, \gamma +\frac{1}{\gamma }.

Ok thanks. I think I know where I stuffed up. :D

Could someone help me with this question? When I form the integral I get: \sqrt{3}\int_{-2}^{2}\sqrt{1-\frac{x^2}{4}} \ dx But in the solutions they don't have a square root on the integrand, I'm not sure what I'm doing wrong :confused:. Thanks. The base of a particular solid is the...

Re: HSC 2015 2U Marathon Good method. But shouldn't the fractions in the first line be a^2/b and b^2/a? I just went: \frac{\alpha ^{2}+\beta ^{2}}{\alpha }+\frac{\alpha ^{2}+\beta ^{2}}{\beta } =\frac{(\alpha +\beta )(\alpha ^{2}+\beta ^{2})}{\alpha \beta }=\frac{(\alpha +\beta )((\alpha...

Re: HSC 2015 2U Marathon Let \ $\alpha \ $and $ \beta \ $be \ the \ roots \ of \ 5x^2+10x+\frac{1}{2}=0, \ find: \frac{\alpha ^{2}+\beta ^{2}}{\alpha }+\frac{\alpha ^{2}+\beta ^{2}}{\beta }

Re: HSC 2015 4U Marathon let equation = P(x) P'(x)=4x^3+6x^2+8 P'(x)=(x+2)(4x^2-2x+4) Therefore x=-2, y=-1 is the only stationary point and thus 2 real roots exist, with one root being between -3 and -2 and the other being between -2 and -1.

Re: HSC 2015 4U Marathon - Advanced Level II. E'(x)=1+x+x^2/2!...+x^(n-1)/(n-1)! At stationary point, E'(x)=0, -1=x+x^2/2!...+x^(n-1)/(n-1)! Therefore at stationary point E(x)=1-1+x^n/n!=x^n/n!, and as n is even, E(x) is positive at any stationary point and as the polynomial is of even degree...

Re: HSC 2015 4U Marathon - Advanced Level I. E(x)-E'(x)=(x^n)/n! Which cannot equal zero. But for a double root, this must equal zero when the root is subbed in. Therefore a double root does not exist. II. I'll try it tomorrow :p

inb4enigma

Re: HSC 2015 4U Marathon $ The statement means that the triangle with sides \left |a \right |, \left | b \right |, \left | a-b \right | $is a right-angled triangle (ratio of sides). \therefore $ \ bcis90^{\circ}=\frac{4a}{3} \Rightarrow 16a^2+9b^2=0 --- $Show that cos^5\alpha...

Re: HSC 2015 4U Marathon $a)$ \ =(3+i)(5+19i)=-4+62i $b)$ \ =\frac{i(27+14i)}{-1325+1059i}=\frac{(-14+27i)(-1325-1059i)}{(-1325+1059i)(-1325-1059i)}=\frac{47143-20949i}{2877106} --- $If \alpha , \beta , \gamma$ \ \ $are the roots of the equation x^3-px^2+qx-r=0, $find \alpha ^3 + \beta ^3...

Re: HSC 2015 4U Marathon $a)$\ (1+i)(3+2i)=3+2i+3i-2=1+5i $b)$\ \frac{2-i}{7 + 17i} =\frac{(2-i)(7-17i)}{(7+17i)(7-17i)}=\frac{14-34i-7i-17}{49+289}=\frac{-3-41i}{338} --- $Show that if \ $x^3+px+r=0$ has a root of multiplicity 2, then \ 27r^2+4p^3=0

Re: HSC 2014 4U Marathon - Advanced Level I agree with this.

Just draw: - a point - then coming off of that point is three branches: W1, W2, B - then coming off of those is two more branches each (for a total of six pathways), with either W1, W2 or B coming off of those (the one you have already used in a pathway doesn't get repeated)

I think the question is asking for the probability of picking a white and a black ball overall, not a white ball then a black ball... is that what you were getting at?

W-W W-W W-B W-B B-W B-W a. 2/6 = 1/3 b. 4/6 = 2/3

Re: HSC 2014 4U Marathon Ok. I wasn't that confident about how I did it anyway :biglaugh: