Stuck on previous exam questions!

Denominator · 2012-03-23 18:13:21

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.

bobbym · 2012-03-23 21:14:58

Hi;

I have a question are a,b,c integers?

Denominator · 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

anonimnystefy · 2012-03-23 21:31:30

Hi Denominator

You should probably not done that. Instead you should have x-2 as a factor of that second polynomial.

Bob · 2012-03-23 21:31:36

hi Denominator,

I'll have a go at these but it'll be in about half an hour if that's OK with you.

No problem whether a, b and c are integers or not.

Back shortly.

Bob

anonimnystefy · 2012-03-23 21:38:25

Hi Denominator

Denominator · 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

Last edited by Denominator (2012-03-23 21:42:17)

bobbym · 2012-03-23 21:42:45

Hi;

anonimnystefy · 2012-03-23 21:46:25

What are those?

bobbym · 2012-03-23 21:49:39

The sums of the squares of the roots, I think.

anonimnystefy · 2012-03-23 21:59:13

Look at mine.

Bob · 2012-03-23 22:05:03

hi Denominator

This is the corrected version.

let the equations be

and

multiplying out

and

equating coefficients gives

from (ii) and (iii)

and putting (ii) into (i) gives

Putting (iv) into (v) gives

which looks a bit nasty. But you can easily spot that R - 1 is a factor which leaves a quadratic, which factorises easily and it should be plain sailing from there.

Bob

ps. Now to read number 2

bobbym · 2012-03-23 22:05:44

Hi anonimnystefy;

Who should be looking?

anonimnystefy · 2012-03-23 22:08:19

You, m.

bobbym · 2012-03-23 22:10:46

There are other solutions if we relax the integer coefficients idea. What are your quadratic two equations?

anonimnystefy · 2012-03-23 22:14:07

They are both 2x^2-5x+2.

Last edited by anonimnystefy (2012-03-23 22:14:34)

bobbym · 2012-03-23 22:15:24

That equation is illegal.

You also should have posted a new thread instead of tacking on to an active one.

Bob · 2012-03-23 22:20:30

hi Denominator

trains:

For questions like this I like to sketch a distance time graph so i get a good picture in my head of what is happening. see below.

You have enough data to work out the gradients for the two lines. And you know that at the moment of passing the 't' values and the 'd' values are equal.

So you could make two equations for d/t based on the gradients and solve for d and t.

Bob

Denominator · 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..

bobbym · 2012-03-24 07:31:29

Hi;

Isn't 5 also an answer?

Denominator · 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

bobbym · 2012-03-24 09:36:31

Why can't you switch the equations calling the second one the original one like this?

Original equation:

Different equation:

Denominator · 2012-03-24 09:40:09

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

bobbym · 2012-03-24 09:44:46

Hi;

Okay, yes you are right, I did not see that.

Bob · 2012-03-24 12:07:49

hi Denominator

Let the distance from Southampton to London be D, and the distance to the overtaking point be d.

Let the time in minutes from 9.17 to the overtaking point be t

The goods train takes 165 minutes
The express train takes 100 minutes

So the gradient (velocity in miles / min) for the goods train is

and for the express train

You can eliminate both D and d by dividing these equations:

and then solve for t.

Bob

Math Is Fun Forum

#1 2012-03-23 18:13:21

Stuck on previous exam questions!

#2 2012-03-23 21:14:58

Re: Stuck on previous exam questions!

#3 2012-03-23 21:27:26

Re: Stuck on previous exam questions!

#4 2012-03-23 21:31:30

Re: Stuck on previous exam questions!

#5 2012-03-23 21:31:36

Re: Stuck on previous exam questions!

#6 2012-03-23 21:38:25

Re: Stuck on previous exam questions!

#7 2012-03-23 21:41:30

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#8 2012-03-23 21:42:45

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#9 2012-03-23 21:46:25

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#10 2012-03-23 21:49:39

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#11 2012-03-23 21:59:13

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#12 2012-03-23 22:05:03

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