Math help (1 Viewer)

[SpiralFlex]

^ I did some work over summer and it helped a bit.

But make sure you know the basics and are solid with it.

 

[RealiseNothing]

^ I did some work over summer and it helped a bit.

But make sure you know the basics and are solid with it.

I did 120 hours of maths in January

(4 hours a day, 30 days in the month).

Helped so much.

 

[SpiralFlex]

I did 120 hours of maths in January

(4 hours a day, 30 days in the month).

Helped so much.

I did two days 5 hours (longest period I have ever concentrated)

 

[Fawun]

inb4 Fawun is best here at maths in 2014

Never lol

In any situation, not only in this sum/product of roots questions, if you have 2 unknowns, you need 2 equations with those 2 unknowns in them to solve the question or at least one equation which solves one of the unknown variables.

The question asking to find k, since you have one equation which solves for k, you only need that equation. Why would you bother with another when you've already answered the question? Now, in the previous equation, you have one equation with a and m as unknown variables. You must have another equation with either 1 or 2 of the variables in order to solve for either of the unknowns. So after you found what 'a' was, you now have 2 equations which can be used to find the value of m.

You need to know what you're trying to solve and look for relevant equations which help you solve for those unknowns.

So if the question is "One root of the quadratic equation x^2 + bx + c = 0 is 4 and the product of roots is 8, Find the values of b and c."

I would have to solve it using two equations right? so I have to solve it using both sum AND products?

Yeah let's not get ahead of ourselves

Nah but seriously- see Fawun- you're improving rapidly. I'm serious, by year 12 you could legit be heaps good at maths. I believe in you

Hehe thanks Deswa

If fawun did what I did in the coming January holidays, she legit could be very good.

Did you get ahead and start year 11 maths or like did year 10 maths?

 

[SpiralFlex]

Basics first.

 

[RealiseNothing]

Did you get ahead and start year 11 maths or like did year 10 maths?

I had HSC maths done before starting year 11.

Although I've forgotten about 10% of the stuff I did which I need to revise.

 

[Fawun]

I had HSC maths done before starting year 11.

Although I've forgotten about 10% of the stuff I did which I need to revise.

??

No wonder you're a genius. Are you accelerating?

 

[HSC2014]

Btw when am I meant to learn all this root stuff? It's all foreign to me. Is it year 10?

 

[RealiseNothing]

??

No wonder you're a genius. Are you accelerating?

Yes I am lol, you wouldn't expect that considering my school would you.

Btw when am I meant to learn all this root stuff? It's all foreign to me. Is it year 10?

It's year 11 I think.

 

[SpiralFlex]

Btw when am I meant to learn all this root stuff? It's all foreign to me. Is it year 10?

Formally learnt in Year 11 (can vary depending on your school) but it is doable for Year 10 level.

 

[Demento1]

Yeah let's not get ahead of ourselves

Nah but seriously- see Fawun- you're improving rapidly. I'm serious, by year 12 you could legit be heaps good at maths. I believe in you

She will overtake me in my maths by 2014.

N.B. She already has because I don't accelerate as much as her - just find it unnecessary. Oh well.

 

[Ealdoon]

Just remember Fawun, you'll always be better at maths than me

 

[Sy123]

Maybe you should have a go at Realise's question again?

http://community.boredofstudies.org/showthread.php?t=291369&p=6101314&viewfull=1#post6101314

 

[Fawun]

Maybe you should have a go at Realise's question again?

http://community.boredofstudies.org/showthread.php?t=291369&p=6101314&viewfull=1#post6101314

Nope. Still can't do it

PART B

I don't understand anything. Legit, for the past 5 questions that they have given that are similar to this, I have basically been copying answers from the back because I don't get anything. The only reason as to why i'm up to part B for this question was because in part A, I just copied what Spiral did except change it to suit this question

 

[Sy123]

Nope. Still can't do it

PART B

I don't understand anything. Legit, for the past 5 questions that they have given that are similar to this, I have basically been copying answers from the back because I don't get anything. The only reason as to why i'm up to part B for this question was because in part A, I just copied what Spiral did except change it to suit this question

You will need to do both sum and product of roots, then solve simultaneously.



Let our roots be alpha and beta, however the condition is that both roots are the same, so therefore beta must equal alpha. Hence our roots are then, alpha and alpha.

Using sum of roots:




So we have one set of equations, we need to find another one that relates alpha and k, and we can get that by product of roots:



Now lets solve simultaneously, so what we need to do is, is that we need to eliminate alpha since we need k since that is what the question is asking for. Looking at equation 1, we see alpha and equation 2 we see alpha squared. So lets square both sides of the first equation in order to get alpha squared IN equation 1.



Lets equate them:

 

[D94]

I don't understand anything. Legit, for the past 5 questions that they have given that are similar to this, I have basically been copying answers from the back because I don't get anything. The only reason as to why i'm up to part B for this question was because in part A, I just copied what Spiral did except change it to suit this question

What does it mean when the "roots are equal"? In this situation, we want something along the lines of (x-a)²=0, so when you solve the equation, you get that x = a and x = a are the solutions. So, the two roots are equal here.

 

[deswa1]

@Sy- k=-1 is also a valid answer because the equation pretty much decomposes to x^2=0 which trivially has two equal roots

 

[Fawun]

Using sum of roots:

Do you always have to do both sum and products? and How come you have two unknowns in the equation? aren't you suppose to have one?

So we have one set of equations, we need to find another one that relates alpha and k, and we can get that by product of roots:

Why don't you find alpha by square rooting the RHS? instead of keeping it as alpha squared?

Now lets solve simultaneously, so what we need to do is, is that we need to eliminate alpha since we need k since that is what the question is asking for. Looking at equation 1, we see alpha and equation 2 we see alpha squared. So lets square both sides of the first equation in order to get alpha squared IN equation 1.

Why alpha squared and not alpha?

What lol. How is the (k+1)² the numerator?

Lets equate them:

How does the whole thing equal to 0... One line it's a fraction with squares and the next line it's 0...

What does it mean when the "roots are equal"? In this situation, we want something along the lines of (x-a)²=0, so when you solve the equation, you get that x = a and x = a are the solutions. So, the two roots are equal here.

Why is it (x-a)²=0 ? I don't get it.

@Sy- k=-1 is also a valid answer because the equation pretty much decomposes to x^2=0 which trivially has two equal roots

How did you get -1? because the answer says that k=0,-1

 

[Sy123]

@Sy- k=-1 is also a valid answer because the equation pretty much decomposes to x^2=0 which trivially has two equal roots

Ah yes, I was a bit too rash in simplifying, turns out I divided by zero when cancelling out.

Real solution:

 

[D94]

Why is it (x-a)²=0 ? I don't get it.

That's a polynomial of degree two, just like your question. This polynomial has two equal roots. It's the same as (x-ß)(x-ß) = 0, where 'a' are the roots. It is in the same 'form' as the polynomial in the question. If you expand it, you get x²-2ßx+ß². (I changed it to beta to avoid confusion)

I'm trying to show you a situation where the 'roots are equal'. You have to know what it means before attempting the question. Do you even know what it means now?

Two equal roots are essentially saying the roots or the solutions of the polynomial are the same/equal/identical.