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#127 Re: Exercises » Dividing Polynomials - Mixed Sums » 2012-08-06 17:01:28

Hello,

*ignores what has just been said and does the actual exercise*

I will the rest later.

#130 Re: Jai Ganesh's Puzzles » 10 second questions » 2012-08-03 15:57:49

Hey

Yeah I must have added an extra zero in my working somewhere XD, sorry!

#133 Re: Jai Ganesh's Puzzles » Oral puzzles » 2012-08-01 19:32:16

Hi guys;

This took me longer than expected to answer >_<

#136 Re: Jai Ganesh's Puzzles » 10 second questions » 2012-04-04 22:30:20

Hey,

Sorry, I don't know how to solve this, D:
Can you tell me the solution?
Thanks

#139 Re: Help Me ! » Stuck on previous exam questions! » 2012-03-26 15:39:03

Yes, I understand this now

Thank you bob and bobby and anonim!

#140 Re: Help Me ! » Stuck on previous exam questions! » 2012-03-24 09:40:09

Because the question says that the different equation has to have 2 as a root

#141 Re: Help Me ! » Stuck on previous exam questions! » 2012-03-24 09:29:24

I don't think so because the question asks for the sum of the roots of the ORIGINAL equation, not the different equation
That is if you're thinking the roots are 2 and 1, but that is from the different equation, not the original

Unless your saying the roots of the original equation are 1/2 and √19 / 2
Which I don't think is right

#142 Re: Help Me ! » Stuck on previous exam questions! » 2012-03-24 07:19:00

Oh okay thanks guys!

Yeah, after using your methods, I get 5/4 as well

I tried doing something different:
Let the solutions of the different equation be X and 2
The solutions of the original equation is 1/X and 1/2 (after playing around, I found this rule)
So 1/X=X
Which means X=1 or -1
Sum of squares (1)²+(1/2)²=5/4

Btw even though 2x²-5x+2 works, the question says 2 different equations, 1 original and the other with the coefficients swapped... this stays the same.
Thanks a lot guys!

As for Q2
Can you tell me what equations you formed, because I couldn't make some..

#143 Re: Help Me ! » Stuck on previous exam questions! » 2012-03-23 21:41:30

Oh okay anonimnystefy,

so I make the equation: (x-2)(ax+b)=0
ax²+(b-2a)x-2b=0
And the other equation is -2bx²+(b-2a)x+a=0
Which factorises to (bx+a)(-2x+1)=0
And the solutions are -a/b and 1/2
Okay so either -b/a=-a/b or -b/a=1/2
Doesn't that mean theres still many solutions for a and b?

Also, I don't have the answers so I'm trusting you guys ;S

#144 Re: Help Me ! » Stuck on previous exam questions! » 2012-03-23 21:27:26

Yeah, probably

I used the quadratic formula to solve for the two equations,
but I don't know which one to be 2 or to be equal with the other roots from the other equation

#145 Help Me ! » Stuck on previous exam questions! » 2012-03-23 18:13:21

Denominator
Replies: 30

Hey guys!

Well I've been trying to do previous exams as practice for the one I have next week.

I came across 2 questions, of which I have no idea how to solve.

If you guys could show working and explanations so that I understand as well, that'll be great.

Q 1
While attempting to solve a quadratic equation, Christobel inadvertently interchanged the coefficient of x² with the constant term, causing the equation to change.
She solved this different equation accurately.
One of the roots she got was 2 and the other was a root of the original equation.
What are the sum of the squares of the two roots of the original equation?

I tried letting the first equation be ax²+bx+c=0 and the other be cx²+bx+a=0 but then I didn't know what to do...

Q 2
A goods train leaves Southampton at 9.17 am and arrives in London at 12.02 pm.
On the same day, a passenger express train leaves Southampton at 9.56 am and arrives in London at 11.36 am.
At what time does the passenger train pass the goods train if each is travelling non-stop at constant speed?

Yeah... I didn't even bother trying this question, I have no clue as to what variables I need to make or such..

Any help is appreciated. Thanks guys.

#146 Re: Jai Ganesh's Puzzles » 5 Star Questions » 2012-03-22 17:50:06

Hey guys,

The ones I answered:

The rest are too hard for me XD

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