Misaki

Thanks for the help

ok thanks

Oh ok thanks!

Ok thanks for the tips, usally i forget them.

And yes by hand

just solve the quadratic equation below, then solve for  a and b

Thanks!

So it would be a divergent series?

Sure! Say you have x^-1 that would be

$1\over x$ if you have x^2 you would get $1\over x^2$ and so on.. but if you have say 

$2x^{-2}$you would get $2 \over x^2$

if you have a fraction that is negative for example $-1\over 2$you would get

$1 \over \sqrt{x}$ , i'm assuming you've already learnt the fraction indice laws.

5.6%*26=1.456

26-1.456=24.544

any side a^2=b^2+c^2-2b*cos(A) for any 

for finding angles $Cos(A)=b^2+c^2-a^2\over 2bc$

What do you mean solve a negative integer?

For question 1

Let $y=x^2-4x$

Let $x=y^2-4y$

With this $x \neq y$

We want to know what $x^2+y^2$ so we square both of the equations to get that

$x^2=x^4-8x^3+16x^2$ and with $y$ we get

$y^2=y^4-8y^3+16y^2$ Since there are differerent coefficents on both $x$ and $y$

All we can do is put it in biggest degree to lowest degree

$x^2+y^2=x^4+y^4−8x^3−8y^3+16x^2+16y^2$

Question 2. $4=a+\frac{1}{a}$ multiply both sides by a 

$4a=a^2+1$ as you notice this a quadratic in disguise! hence

$-a^2+4a-1=0$ as factorisation dosn't work we will use the quadratic formula

For values a=-1  b=4 c=-1 

$a = {-4 \pm \sqrt{4^2-4} \over -2}$simplify $a = {2 \pm \sqrt{12} }$

Simplify even more $a=2 \pm 2\sqrt{3}$

Now to find the value of $a^4+a^{-4}$$a^4=97\pm56\sqrt{3}$$a^{-4}=\frac{7}{64}\pm \frac{1}{16}\sqrt{3}$

$a^4 + a^{-4}= 97 \pm 56\sqrt{3} + \frac{7}{64} \pm \frac{1}{16}\sqrt{3}$ after a bit more simplifying we get   

$\frac{6215}{64}-\frac{897}{16}\sqrt{3}$ if $97 - 56\sqrt{3} + \frac{7}{64} - \frac{1}{16}\sqrt{3}$ if $97 + 56\sqrt{3} + \frac{7}{64} + \frac{1}{16}\sqrt{3}$ we get $\frac{6215}{64}+\frac{897}{16}\sqrt{3}$ if $97 + 56\sqrt{3} + \frac{7}{64} - \frac{1}{16}\sqrt{3}$ we get$\frac{6215}{64}+\frac{895}{16}\sqrt{3}$ if

$97 - 56\sqrt{3} + \frac{7}{64} + \frac{1}{16}\sqrt{3}$ we get $\frac{6215}{64}-\frac{895}{16}\sqrt{3}$ that's all the combinations! PHEW...

It's ok i still understood you!

Thank you very much! 

you need to know the height for both of the formula's