Comments on: The Root Cause Of Miss Loi’s Virus http://www.exampaper.com.sg/questions/a-maths/the-root-cause-of-miss-lois-virus Sassy O Level Maths Tuition, Questions & Tips from Singapore's Favourite Private Tutor Sun, 05 Feb 2012 17:08:39 +0000 hourly 1 By: Miss Loi http://www.exampaper.com.sg/questions/a-maths/the-root-cause-of-miss-lois-virus#comment-14007 Miss Loi Mon, 29 Sep 2008 05:15:37 +0000 http://www.exampaper.com.sg/questions/a-maths/the-root-cause-of-miss-lois-virus#comment-14007 And so it came to pass that on a historic weekend when <a href="/miss-loi-temple" rel="nofollow">The Temple</a> reverberated with the sounds of <a rel="external nofollow" href="http://www.singaporegp.sg" rel="nofollow">roaring F1 cars</a>, Miss Loi was finally healed by a potent vaccine that was concocted by <b>Someone</b>, using ingredients from an <em>A-Level</em> 武林密籍 that overwhelmingly killed the virus in an instant. P.S. O-Level students are advised not to be alarmed by the <i>chim</i> concepts presented in <a href="/questions/a-maths/the-root-cause-of-miss-lois-virus#comment-13976" rel="nofollow">Comment #9</a>, for you'll ONLY be dealing with <em>two</em> roots in your exam - so just focus on the stuff in the orange boxes within the main blog post. P.P.S. The typo mentioned in <a href="/questions/a-maths/the-root-cause-of-miss-lois-virus#comment-13923" rel="nofollow">Comment #6</a> has been fixed. Miss Loi was <em>really</em> sick when she put out this blog post (yeah yeah excuses excuses ...) Last but not least, thanks <b>FoxTwo</b> and <b>Daniel</b> for pats pats and your well-wishes. The sneezing has completely stopped now :D And so it came to pass that on a historic weekend when The Temple reverberated with the sounds of roaring F1 cars, Miss Loi was finally healed by a potent vaccine that was concocted by Someone, using ingredients from an A-Level 武林密籍 that overwhelmingly killed the virus in an instant.

P.S. O-Level students are advised not to be alarmed by the chim concepts presented in Comment #9, for you'll ONLY be dealing with two roots in your exam - so just focus on the stuff in the orange boxes within the main blog post.

P.P.S. The typo mentioned in Comment #6 has been fixed. Miss Loi was really sick when she put out this blog post (yeah yeah excuses excuses ...)

Last but not least, thanks FoxTwo and Daniel for pats pats and your well-wishes. The sneezing has completely stopped now :D

]]> By: Someone http://www.exampaper.com.sg/questions/a-maths/the-root-cause-of-miss-lois-virus#comment-13976 Someone Sun, 28 Sep 2008 06:37:38 +0000 http://www.exampaper.com.sg/questions/a-maths/the-root-cause-of-miss-lois-virus#comment-13976 I prefer to use [tex]{\sum}{\alpha}=-{\frac{b}{a}}[/tex] [tex]{\sum}{\alpha}{\beta}={\frac{c}{a}}[/tex] [tex]{\sum}{\alpha}{\beta}{\gamma}=-{\frac{d}{a}}[/tex] [tex]{\sum}{\alpha}{\beta}{\gamma}{\delta}={\frac{e}{a}}[/tex] because it helps me to remember the general form of Vieta's forumulas when you sum [tex]\alpha[/tex], you take the next letter b (taking alpha as the letter a), with a minus sign in front BECAUSE its an ODD number of roots, getting [tex]-{\frac{b}{a}}[/tex] when you sum [tex]\alpha\beta[/tex], you take the next letter c (taking beta as letter b), without any minus sign in front BECAUSE its an EVEN number of roots getting [tex]{\frac{c}{a}}[/tex] I prefer to use
{\sum}{\alpha}=-{\frac{b}{a}}
{\sum}{\alpha}{\beta}={\frac{c}{a}}
{\sum}{\alpha}{\beta}{\gamma}=-{\frac{d}{a}}
{\sum}{\alpha}{\beta}{\gamma}{\delta}={\frac{e}{a}}
because it helps me to remember the general form of Vieta's forumulas

when you sum \alpha, you take the next letter b (taking alpha as the letter a), with a minus sign in front BECAUSE its an ODD number of roots, getting -{\frac{b}{a}}

when you sum \alpha\beta, you take the next letter c (taking beta as letter b), without any minus sign in front BECAUSE its an EVEN number of roots getting {\frac{c}{a}}

]]> By: Someone http://www.exampaper.com.sg/questions/a-maths/the-root-cause-of-miss-lois-virus#comment-13975 Someone Sun, 28 Sep 2008 06:28:04 +0000 http://www.exampaper.com.sg/questions/a-maths/the-root-cause-of-miss-lois-virus#comment-13975 Haha. My choice of notation is from the all powerful Bostock and Chandler. Heehee. Yay! I got it correct. Woot. Haha. My choice of notation is from the all powerful Bostock and Chandler. Heehee. Yay! I got it correct. Woot.

]]> By: Daniel http://www.exampaper.com.sg/questions/a-maths/the-root-cause-of-miss-lois-virus#comment-13961 Daniel Sat, 27 Sep 2008 18:02:22 +0000 http://www.exampaper.com.sg/questions/a-maths/the-root-cause-of-miss-lois-virus#comment-13961 Hi Ms Loi, take care and get well soon! :) Hi Ms Loi, take care and get well soon! :)

]]> By: Someone http://www.exampaper.com.sg/questions/a-maths/the-root-cause-of-miss-lois-virus#comment-13923 Someone Fri, 26 Sep 2008 17:28:43 +0000 http://www.exampaper.com.sg/questions/a-maths/the-root-cause-of-miss-lois-virus#comment-13923 By the way, shouldn't: and the above equation can be rewritten as: [tex]{{x}^2}+(sum)x+(product)=0[/tex] be: and the above equation can be rewritten as: [tex]{{x}^2}-(sum)x+(product)=0[/tex] since [tex]sum=-\frac{b}{a}[/tex] By the way, shouldn't:
and the above equation can be rewritten as:
{{x}^2}+(sum)x+(product)=0

be:

and the above equation can be rewritten as:
{{x}^2}-(sum)x+(product)=0

since sum=-\frac{b}{a}

]]> By: Someone http://www.exampaper.com.sg/questions/a-maths/the-root-cause-of-miss-lois-virus#comment-13922 Someone Fri, 26 Sep 2008 17:23:29 +0000 http://www.exampaper.com.sg/questions/a-maths/the-root-cause-of-miss-lois-virus#comment-13922 <span class="highlight"> * Miss Loi takes a break from her sneezing * </span> Part 1 [tex]4{x^2} +(-1) x + 36=0[/tex] [tex]{$\alpha$}^{\prime}={$\alpha$}^{2}[/tex] [tex]{$\beta$}^{\prime}={$\beta$}^{2}[/tex] [tex]{\sum}{$\alpha$}^{\prime}=-\frac{b}{a}=-\frac{-1}{4}=\frac{1}{4}[/tex] [tex]{\sum}{$\alpha$}^{\prime}{$\beta$}^{\prime}=\frac{c}{a}=\frac{36}{4}=9[/tex] [tex]\frac{1}{{$\alpha$}^{2}}+\frac{1}{{$\beta$}^{2}}=\frac{1}{{$\alpha$}^{\prime}}+\frac{1}{{$\beta$}^{\prime}}=\frac{{$\beta$}^{\prime}}{{$\alpha$}^{\prime}{$\beta$}^{\prime}}+\frac{{$\alpha$}^{\prime}}{{$\beta$}^{\prime}{$\alpha$}^{\prime}}=\frac{{$\beta$}^{\prime}+{$\alpha$}^{\prime}}{{$\alpha$}^{\prime}{$\beta$}^{\prime}}=\frac{\frac{1}{4}}{9}=\frac{({\frac{1}{4}})(4)}{(9)(4)}=\frac{1}{36}[/tex] [tex](\frac{1}{{$\alpha$}^{2}})(\frac{1}{{$\beta$}^{2}})=(\frac{1}{{$\alpha$}^{\prime}})(\frac{1}{{$\beta$}^{\prime}})=\frac{1}{{$\alpha$}^{\prime}{$\beta$}^{\prime}}=\frac{1}{9}[/tex] [tex]-\frac{b^{\prime}}{a^{\prime}}=\frac{1}{36}[/tex] [tex]\frac{c^{\prime}}{a^{\prime}}=\frac{1}{9}[/tex] [tex]{{x}^2}-({-\frac{b^{\prime}}{a^{\prime}}){x}+{\frac{c^{\prime}}{a^{\prime}}={{x}^2}-{\frac{1}{36}}{x}+\frac{1}{9}[/tex] The equation is: [tex]{{x}^2}-{\frac{1}{36}}{x}+\frac{1}{9}=0[/tex] <div class="highlight"> It's interesting that you've chosen to use additional notations i.e. letting <var>α</var>' = <var>α</var><sup>2</sup>, Σ<var>α</var>' etc. etc. But to keep to our O-Level scope, we're dealing with <em>quadratic</em> roots (not higher order polynomials) so we're pretty sure that there'll only be <em>two</em> lonely roots <var>α</var> & <var>β</var> (somehow this reminds me of Wall-E & Eve :) ) So for 4<var>x</var><sup>2</sup> - <var>x</var> + 36 = 0 with roots <var>α</var><sup>2</sup>, <var>β</var><sup>2</sup>: Sum of roots: <m>alpha^2 + beta^2 = 1/4</m> ----- (1) (∵ <m>-b/a</m>) Product of roots: <m>alpha^2 beta^2 = 36/4 = 9</m> ----- (2) (∵ <m>c/a</m>) For an equation with roots <m>1/alpha^2, 1/beta^2</m>: Sum of roots: <m>1/{alpha^2} + 1/{beta^2} = {alpha^2 + beta^2}/{alpha^2 beta^2} = {1/4}/9 = 1/36</m> using (1), (2) above Product of roots: <m>1/{alpha^2 beta^2} = 1/9</m> using (2) above So the equation for part 1 is <m>x^2 - {1/36}x + {1/9} = 0</m> which can also be expressed as 36<var>x</var><sup>2</sup> - <var>x</var> + 4 = 0 ;) P.S. <b>Someone</b>: These are essentially the same steps you've done but just curious on your choice of notation. </div> Part 2 [tex]{{\alpha}^2}{{\beta}^2}=9[/tex] [tex]({\alpha}{\beta})^2=9[/tex] [tex]{\alpha}{\beta}={\pm}\sqrt{9}={\pm}{3}[/tex] [tex]{{({\alpha}+{\beta}})^2}={{\alpha}^2}+{{\beta}^2}+2{{\alpha}{\beta}}[/tex] <div class="highlight"> This is key part of the working. In their haste, many students have forgotten about this expression α<sup>2</sup> + β<sup>2</sup> = (α + β)<sup>2</sup> - 2αβ when trying to calculate α, β from the given α<sup>2</sup>, β<sup>2</sup> and vice versa! </div> [tex]{{({\alpha}+{\beta}})^2}={\frac{1}{4}}+2({\pm}3)={\frac{1}{4}}{\pm}6[/tex] [tex]{{({\alpha}+{\beta}})^2}={\frac{1}{4}}+6=\frac{25}{4}[/tex] [tex]{{({\alpha}+{\beta}})^2}={\frac{1}{4}}-6[/tex] (reject) <span class="highlight">Actually you already know at this point that <var>αβ</var> cannot be negative - since <m>{1/4} - 6</m> is obtained from <var>αβ</var> = -3 ;)</span> [tex]{\alpha}+{\beta}={\pm}\sqrt{\frac{25}{4}}={\pm}\frac{5}{2}[/tex] [tex]{\alpha}+{\beta}={\pm}\frac{5}{2}[/tex] [tex]{\alpha}{\beta}=3[/tex] because when [tex]{\alpha}{\beta}=-3[/tex] , [tex]{{({\alpha}+{\beta}})^2}[/tex] is negative [tex]{{x}^2}-({\pm}\frac{5}{2}){x}+3=0[/tex] The 2 equations are: [tex]{{x}^2}-\frac{5}{2}{x}+3=0[/tex] [tex]{{x}^2}+\frac{5}{2}{x}+3=0[/tex] <span class="highlight">Once again this can also be expressed as: 2<var>x</var><sup>2</sup> ± 5<var>x</var> + 6 = 0 ;)</span> Pardon me if I have any errors, I'm doing this late at night, and with NO PAPER, using latex as working. Haha * Miss Loi takes a break from her sneezing *

Part 1
4{x^2} +(-1) x + 36=0
{$\alpha$}^{\prime}={$\alpha$}^{2}
{$\beta$}^{\prime}={$\beta$}^{2}
{\sum}{$\alpha$}^{\prime}=-\frac{b}{a}=-\frac{-1}{4}=\frac{1}{4}
{\sum}{$\alpha$}^{\prime}{$\beta$}^{\prime}=\frac{c}{a}=\frac{36}{4}=9
\frac{1}{{$\alpha$}^{2}}+\frac{1}{{$\beta$}^{2}}=\frac{1}{{$\alpha$}^{\prime}}+\frac{1}{{$\beta$}^{\prime}}=\frac{{$\beta$}^{\prime}}{{$\alpha$}^{\prime}{$\beta$}^{\prime}}+\frac{{$\alpha$}^{\prime}}{{$\beta$}^{\prime}{$\alpha$}^{\prime}}=\frac{{$\beta$}^{\prime}+{$\alpha$}^{\prime}}{{$\alpha$}^{\prime}{$\beta$}^{\prime}}=\frac{\frac{1}{4}}{9}=\frac{({\frac{1}{4}})(4)}{(9)(4)}=\frac{1}{36}
(\frac{1}{{$\alpha$}^{2}})(\frac{1}{{$\beta$}^{2}})=(\frac{1}{{$\alpha$}^{\prime}})(\frac{1}{{$\beta$}^{\prime}})=\frac{1}{{$\alpha$}^{\prime}{$\beta$}^{\prime}}=\frac{1}{9}
-\frac{b^{\prime}}{a^{\prime}}=\frac{1}{36}
\frac{c^{\prime}}{a^{\prime}}=\frac{1}{9}
{{x}^2}-({-\frac{b^{\prime}}{a^{\prime}}){x}+{\frac{c^{\prime}}{a^{\prime}}={{x}^2}-{\frac{1}{36}}{x}+\frac{1}{9}
The equation is:
{{x}^2}-{\frac{1}{36}}{x}+\frac{1}{9}=0

It's interesting that you've chosen to use additional notations i.e. letting α' = α2, Σα' etc. etc.

But to keep to our O-Level scope, we're dealing with quadratic roots (not higher order polynomials) so we're pretty sure that there'll only be two lonely roots α & β (somehow this reminds me of Wall-E & Eve :) )

So for 4x2 - x + 36 = 0 with roots α2, β2:

Sum of roots: alpha^2 + beta^2 = 1/4 ----- (1) (∵ -b/a)
Product of roots: alpha^2 beta^2 = 36/4 = 9 ----- (2) (∵ c/a)

For an equation with roots 1/alpha^2, 1/beta^2:

Sum of roots:
1/{alpha^2} + 1/{beta^2} = {alpha^2 + beta^2}/{alpha^2 beta^2} = {1/4}/9 = 1/36 using (1), (2) above
Product of roots: 1/{alpha^2 beta^2} = 1/9 using (2) above

So the equation for part 1 is x^2 - {1/36}x + {1/9} = 0
which can also be expressed as 36x2 - x + 4 = 0 ;)

P.S. Someone: These are essentially the same steps you've done but just curious on your choice of notation.

Part 2
{{\alpha}^2}{{\beta}^2}=9
({\alpha}{\beta})^2=9
{\alpha}{\beta}={\pm}\sqrt{9}={\pm}{3}
{{({\alpha}+{\beta}})^2}={{\alpha}^2}+{{\beta}^2}+2{{\alpha}{\beta}}

This is key part of the working. In their haste, many students have forgotten about this expression α2 + β2 = (α + β)2 - 2αβ when trying to calculate α, β from the given α2, β2 and vice versa!

{{({\alpha}+{\beta}})^2}={\frac{1}{4}}+2({\pm}3)={\frac{1}{4}}{\pm}6
{{({\alpha}+{\beta}})^2}={\frac{1}{4}}+6=\frac{25}{4}
{{({\alpha}+{\beta}})^2}={\frac{1}{4}}-6 (reject)

Actually you already know at this point that αβ cannot be negative - since {1/4} - 6 is obtained from αβ = -3 ;)

{\alpha}+{\beta}={\pm}\sqrt{\frac{25}{4}}={\pm}\frac{5}{2}
{\alpha}+{\beta}={\pm}\frac{5}{2}
{\alpha}{\beta}=3 because when {\alpha}{\beta}=-3 , {{({\alpha}+{\beta}})^2} is negative

{{x}^2}-({\pm}\frac{5}{2}){x}+3=0

The 2 equations are:
{{x}^2}-\frac{5}{2}{x}+3=0
{{x}^2}+\frac{5}{2}{x}+3=0

Once again this can also be expressed as: 2x2 ± 5x + 6 = 0 ;)

Pardon me if I have any errors, I'm doing this late at night, and with NO PAPER, using latex as working. Haha

]]> By: FoxTwo http://www.exampaper.com.sg/questions/a-maths/the-root-cause-of-miss-lois-virus#comment-13871 FoxTwo Thu, 25 Sep 2008 16:33:06 +0000 http://www.exampaper.com.sg/questions/a-maths/the-root-cause-of-miss-lois-virus#comment-13871 Awww *pat pat* come come I make chicken soup for you.... Awww *pat pat* come come I make chicken soup for you....

]]> By: Miss Loi http://www.exampaper.com.sg/questions/a-maths/the-root-cause-of-miss-lois-virus#comment-13870 Miss Loi Thu, 25 Sep 2008 16:10:19 +0000 http://www.exampaper.com.sg/questions/a-maths/the-root-cause-of-miss-lois-virus#comment-13870 <b>FoxTwo:</b> ... *sneeze* <b>Piak:</b> Thanks for your fervent support *sneeze* FoxTwo: ... *sneeze*

Piak: Thanks for your fervent support *sneeze*

]]> By: piakpiak http://www.exampaper.com.sg/questions/a-maths/the-root-cause-of-miss-lois-virus#comment-13847 piakpiak Wed, 24 Sep 2008 18:43:21 +0000 http://www.exampaper.com.sg/questions/a-maths/the-root-cause-of-miss-lois-virus#comment-13847 永远支持你。miss loi,miss loi number 1 永远支持你。miss loi,miss loi number 1

]]> By: FoxTwo http://www.exampaper.com.sg/questions/a-maths/the-root-cause-of-miss-lois-virus#comment-13843 FoxTwo Wed, 24 Sep 2008 15:33:56 +0000 http://www.exampaper.com.sg/questions/a-maths/the-root-cause-of-miss-lois-virus#comment-13843 Umm.... you *did* know I was kidding when I suggested you make a maths problem out of your recent bout with the flu on Plurk, right? Right? right? Umm.... you *did* know I was kidding when I suggested you make a maths problem out of your recent bout with the flu on Plurk, right?

Right?

right?

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