Demento's Year 10 maths practice paper thread (2 Viewers)

SpiralFlex

Well-Known Member
Joined
Dec 18, 2010
Messages
6,960
Gender
Female
HSC
N/A
I really cbf'd getting them to prove it right now. I started making that question at 9:50, finished it at 11:35 or something. Just relaxing now lol.
Meanie, that's like pushing a child into a pool of sharks.
 

HSC2014

Member
Joined
Jul 30, 2012
Messages
399
Gender
Male
HSC
N/A
It was a test. How can I check the solutions?! lol
I was talking about your solutions to the abs value questions. If you did them by case definitions, then you have to check their met the conditions (If you don't get what I'm saying dw. And there was only 20 questions all together so probably wasn't too many abs values).
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Introduction to Realise's question:

i) Given that




and



Prove that:







Then continue with realise's question.
 
Last edited:

Demento1

Philosopher.
Joined
Dec 23, 2011
Messages
866
Location
Sydney
Gender
Male
HSC
2014
Introduction to Realise's question:

i) Given that



and



Prove that:






Then continue with realise's question.
Will give this a go tomorrow - looking forward to it. Goodnight guys (or morning).
 

Fawun

Queen
Joined
Jun 23, 2012
Messages
1,270
Gender
Undisclosed
HSC
2014
Introduction to Realise's question:

i) Given that



and



Prove that:






Then continue with realise's question.
What the heck how does ax^2 + bx + c turn out to be (x-a)(x-b)

:S
 

HSC2014

Member
Joined
Jul 30, 2012
Messages
399
Gender
Male
HSC
N/A
SpiralFlex, aren't you meant to be studying 24/7 for HSC t_t
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Defining alpha and beta as the roots of course.
Oh yes, of couse lol.

What the heck how does ax^2 + bx + c turn out to be (x-a)(x-b)

:S
well ax^2+bx+c=0.

Do something to the x^2 term (and hence the whole expression) to get it into a desirable form in order to then utilise the equivalence to (x-alpha)(x-beta)
 

SpiralFlex

Well-Known Member
Joined
Dec 18, 2010
Messages
6,960
Gender
Female
HSC
N/A
So no one is going to explain to Fawn? If not I have to then...(eating my pie)
 

Fawun

Queen
Joined
Jun 23, 2012
Messages
1,270
Gender
Undisclosed
HSC
2014
Oh yes, of couse lol.



well ax^2+bx+c=0.

Do something to the x^2 term (and hence the whole expression) to get it into a desirable form in order to then utilise the equivalence to (x-alpha)(x-beta)
idgi

but maybe that's because it's like 12:30am and my brain isn't functioning properly



When you expand the brackets.....
Yes because the LHS looks the same as the RHS
 
Joined
Apr 1, 2011
Messages
1,012
Location
District 12
Gender
Male
HSC
2013
Yes because the LHS looks the same as the RHS
they don't have to, that's why it says "equivalent to" rather than the usual equals sign. We're just establishing the fact that they are identical to each other if that makes sense

(sorry, not the best teacher out there :()
 
Last edited:

SpiralFlex

Well-Known Member
Joined
Dec 18, 2010
Messages
6,960
Gender
Female
HSC
N/A
When we have a univariate equation of a quadratic we are often to find these roots of the equation, either by factorisation and using null factor law or by the quadratic formula.

For example,







We call 2, 3 the roots (sometimes you hear the word zeros, used interchangeably) of the equation.



We will now define the roots to be alpha and beta. There are only two roots, why? Because this is derived from The Fundamental Theorem of Algebra which is

Every non-zero univariable polynomial with complex coefficients has exactly as many complex roots as its degree.

In simpler terms, a quadratic (degree two) has two complex roots. A cubic has three complex roots. But you don't learn the word "complex roots" until later in your studies. So for now we will agree that some quadratics will have two real roots.


Following the example above,

Let's try the sum of the roots,



We see we come up with a value of 5. Let's inspect the polynomial,




Doesn't this look like the second term but with a minus sign?



Let's try the product of the roots,




Let's inspect our polynomial again,



Doesn't this value of 6 agree with the constant term of the polynomial?



Let's see why.


We defined alpha, beta to be the roots of some quadratic equation say,







We can also write it in a products of sums form,






If these two expressions are equivalent, we can equate them. So -










By equating coefficients of x and the constants,



Sum of the roots:



Or sometimes written as,






Product of the roots,



Or sometimes written as




Alternatively,

Consider the quadratic equation,




We can find the roots by using the quadratic equation formula,





Of course alpha, beta are arbitrarily defined so they can be interchangeable.


Sum of roots,








Product of roots:



Using difference between squares in numerator,





 
Last edited:

Demento1

Philosopher.
Joined
Dec 23, 2011
Messages
866
Location
Sydney
Gender
Male
HSC
2014
Proved the sum and product of roots using quadratic equation, based on spiral's answers, I think I am correct.

For my maths paper, I will post solutions once I can get somebody to combine all my PDF files into one to share onto a website link. Give me about a day and you can mark your papers according to my solutions and what I say deserve or doesn't deserve marks.
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Also I will add to Spiral's explanation.

Notice



The fact is, all this time when you were evaluating quadratic equations by using the rule: 'Find two numbers which when we add them you get the middle term and when we multiply them we get the constant'

Like when we factorise:



Lets find two numbers, which when you add them you get 5, when you times them you get 6. Its 2 and 3 right? So our factorised form is:



And hence our roots/zeroes/solutions will be -2 and -3.

Plug it into our formula

So -b/a should be -(-2-3)=5 which it is
and c/a should be 2*3=6 which it is

So I thought it would be useful to note that whenever you are actually doing this method of factorising quadratics, the heart of what you are doing is the sum and product of roots =)
 

Users Who Are Viewing This Thread (Users: 0, Guests: 2)

Top