/sci/ - Science & Math

[Anonymous No.1783656]

Why is 2/(√5-1) ,(√5+1)/2

Also could you explain how its done,many thanks

[Anonymous No.1783663]

2/(√5-1)=2(√5+1)/(√5-1)(√5+1)=2(√5+1)/5-1=(√5+1)/2

[Anonymous No.1783666]

quadratic expansion

[Anonymous No.1783694]

>>1783663
I dont get what you mean

[Anonymous No.1783704]

>>1783694
why?

[Anonymous No.1783708]

BeastUK was such a fucking douchebag. He is the one that troll tipped me over that which convinced me to stop lurking /b/

He's a bald white supremecist fat & pasty limey faggot who thinks way too highly of himself.

I hope he gets raped by a pack of pakistani immigrants.

[Adrian !wf/oOO.fjg No.1783715]

<div class="math">\frac{2}{\sqrt{5}-1} = \frac{2(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)} = \frac{2(\sqrt{5}+1)}{\sqrt{5}\sqrt{5}-\sqrt{5}+\sqrt{5}-1} = \frac{2(\sqrt{5}+1)}{5-1}= \frac{2(\sqrt{5}+1)}{4)= \frac{(\sqrt{5}+1)}{2}</div>

[Anonymous No.1783716]

Pic related,

i dont understand the steps

[Adrian !wf/oOO.fjg No.1783720]

<div class="math"> \frac{2}{\sqrt{5}-1} = \frac{2(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)} = \frac{2(\sqrt{5}+1)}{\sqrt{5}\sqrt{5}-\sqrt{5}+\sqrt{5}-1} = \frac{2(\sqrt{5}+1)}{5-1}= \frac{2(\sqrt{5}+1)}{4})= \frac{(\sqrt{5}+1)}{2}</div>

[Anonymous No.1783726]

$\frac{a}{b+c}=\frac{a(b-c)}{(b+c)(b-c)}\\frac{a(b-c)}{b^2-c^2}

[Anonymous No.1783736]

>>1783720
>>1783663

Big thanks for the help i really appreciate it.

[Anonymous No.1783741]

>>1783726
how to use jsmath?

[Anonymous No.1783755]

$\frac{a}{b+c}=\frac{a(b-c)}{(b+c)(b-c)}\\frac{a(b-c)}{b^2-c^2}$

[Anonymous No.1783819]

how would i do something like √8/(√2-1) then? times both by (√2-1)?