C2 - Logarithms

Original post by Chittesh14

Ok. But, why on a thread about logarithms lol? It doesn't involve logs.
No one will even look at this thread unless they are here to help me or are learning logs too....

It was the most active thread, it's all dead since exams have come and gone,

Reply 62

OP

Original post by Ano123

It was the most active thread, it's all dead since exams have come and gone,

Lol. The Maths section is active - just for A-levels because all GCSE exams have finished. Just wait till next year when you go to A-levels lol, it'll be much more fun as you will be able to relate to the threads too. It's just too active at A-level - mainly because it's tough lool.

Reply 63

Original post by Chittesh14

Lol. The Maths section is active - just for A-levels because all GCSE exams have finished. Just wait till next year when you go to A-levels lol, it'll be much more fun as you will be able to relate to the threads too. It's just too active at A-level - mainly because it's tough lool.

I don't follow what you mean. In what world is this a GCSE question?

Reply 64

OP

Original post by Ano123

I don't follow what you mean. In what world is this a GCSE question?

I never said it was lol?

Reply 65

Original post by Chittesh14

I never said it was lol?

I don't get what you were saying. You said wait next year until you get to A-levels.

Reply 66

OP

Original post by Ano123

I don't get what you were saying. You said wait next year until you get to A-levels.

Aren't you in Year 11 or just finished GCSEs?

Reply 67

Original post by Chittesh14

Aren't you in Year 11 or just finished GCSEs?

Yes, year 11, but doing A-levels.

Reply 68

OP

Original post by Ano123

Yes, year 11, but doing A-levels.

Lol what?
Oh, as in just A-level Maths and just FM GCSE with it?

Reply 69

Original post by Chittesh14

Lol what?
Oh, as in just A-level Maths and just FM GCSE with it?

I'm not doing GCSE further maths.

Reply 70

OP

Original post by Ano123

I'm not doing GCSE further maths.

So you just post questions on the thread for people to attempt?
What are you actually doing lol :s-smilie:

?

Reply 71

I'm a troll in a sense. I just put up questions and stuff I find interesting, let others who might be interested have a go. Look up the hard integral thread, I do what others do in that thread.

Reply 72

SalazarSlytherin

You are wrong, book is right

Reply 73

OP

Original post by Ano123

I'm a troll in a sense. I just put up questions and stuff I find interesting, let others who might be interested have a go. Look up the hard integral thread, I do what others do in that thread.

Lol ok. What A-level modules have you done so far?

Reply 74

Original post by chittesh14

lol ok. What a-level modules have you done so far?

c1-4, m1-2, fp1-4, s1-2

Reply 75

OP

Original post by Ano123

c1-4, m1-2, fp1-4, s1-2

Woah lool. You must do some intense revision. Well done, you must've taken some of those modules as exams too?

Reply 76

Original post by Chittesh14

Woah lool. You must do some intense revision. Well done, you must've taken some of those modules as exams too?

All of them.

Reply 77

OP

Original post by Ano123

All of them.

How did you do in them?

Original post by Ano123

Find the real roots of the equation

\displaystyle x^6+8x^5+16x^4+18x^3+16x^2+8x+1=0

.

It's 10pm, I'm tired, and I really hate polynomials with degree n>3 so usually I'd tell you to go **** yourself... but I gave it a shot anyway with a nice long method, enjoy. =)

Let

f(x)=x^6+8x^5+16x^4+18x^3+16x^2+8x+1

\rightarrow f(-1)=0

therefore

x=-1

is a solution.

---------------------------------------------------------------------------------

\Rightarrow f(x)=0

therefore

(x^3+x^{-3})+8(x^2+x^{-2})+16(x+x^{-1})+18=0

(x+x^{-1})^3=x^3+x^{-3}+3(x+x^{-1}) \Rightarrow (x^3+x^{-3})=(x+x^{-1})^3-3(x+x^{-1})

(x+x^{-1})^2=x^2+x^{-2}+2 \Rightarrow (x^2+x^{-2})=(x+x^{-1})^2-2

\Longrightarrow [(x+x^{-1})^3-3(x+x^{-1})]+8[(x+x^{-1})^2-2]+16(x+x^{-1})+18=0

\Longrightarrow (x+x^{-1})^3+8(x+x^{-1})^2+13(x+x^{-1})+2=0

Let

a=x+x^{-1}

\Rightarrow a^3+8a^2+13a+2=0

Consider roots of the cubic:

\sum(\alpha)=-8, \sum(\alpha\beta)=13, \alpha\beta\gamma=-2

We know

x=-1

is a root for

f(x)=0

therefore

a=(-1)+(-1)^{-1}=-2

therefore

\alpha=-2

Coefficients of cubic are real therefore if

\alpha

is real then

\beta

and

\gamma

are complex conjugates of each other.

\sum(\alpha)=\alpha+\beta+\gamma=-8

therefore

\beta+\gamma=-6

\alpha\beta\gamma=-2

therefore

\gamma=\frac{1}{\beta}

\Rightarrow \beta+\frac{1}{\beta}=-6 \Rightarrow \gamma^2+6\gamma+1=0 \Rightarrow (\gamma+3)^2=8 \Rightarrow \gamma=-3+\sqrt8

therefore

\beta=-3-\sqrt8

Therefore roots of

a^3+8a^2+13a+2=0

are

a=-2 / -3+\sqrt8 / -3-\sqrt8

---------------------------------------------------------------------------------

Coming back to the original

f(x)

...

For

a=-2

;

x+x^{-1}=-2 \Rightarrow x^2+2x+1=0 \Rightarrow (x+1)^2=0 \Rightarrow x=-1

For

a=-3+\sqrt8

;

x+x^{-1}=-3+\sqrt8 \Rightarrow x^2+(3-\sqrt8)x+1=0 \Rightarrow b^2-4ac<0

therefore no real roots.

For

a=-3-\sqrt8

;

x+x^{-1}=-3-\sqrt8 \Rightarrow x^2+(3+\sqrt8)x+1=0 \Rightarrow b^2-4ac=13+12\sqrt2 \Rightarrow x=\frac{-(3+\sqrt8)\pm\sqrt((13+12\sqrt2))}{2}

-----------------------------------------------------------------------------------

Therefore the real solutions to

f(x)=0

are:

x=-1

x= -\frac{(3+\sqrt8)+\sqrt{13+\sqrt2\cdot12}}{2}

x= -\frac{(3+\sqrt8)-\sqrt{13+\sqrt2\cdot12}}{2}

(edited 7 years ago)

Reply 79