/sci/ - Science & Math

[Anonymous No.3713658]

what rule do you use to go from

sqrt(1-x^2/16)
to (1/4)(sqrt(16-x^2)

????
http://www.cramster.com/solution/solution/988167

what the hell did this guy do

[Anonymous No.3713667]

bampies

[Anonymous No.3713682]

I'll give you a big hint, 1 can be rewritten many ways and a fraction in a square root can be split into two square roots.

[Anonymous No.3713683]

Multiply the inside of the sqrt by 16/16. This is valid because it's just multiplying by 1. That gives sqrt((16-x^2)/16). Now you can split that into sqrt(16-x^2)*sqrt(1/16) and that second sqrt gives you the 1/4

[Anonymous No.3713687]

>x^2/16

Confirmed for idiot.

[Anonymous No.3713721]

>>3713683
it took me a while to get what you're saying but now i get it. thanks man

[Anonymous No.3713732]

<span class="math">\sqrt{1-\frac{x^{2}}{16}}[/spoiler]
<span class="math">\sqrt{\frac{16}{16}-\frac{x^{2}}{16}}[/spoiler]
<span class="math">\sqrt{\frac{16-x^{2}}{16}}[/spoiler]
<span class="math">\sqrt{\frac{1}{16}\;16-x^{2}}[/spoiler]
<span class="math">\sqrt{\frac{1}{16}}\cdot\sqrt{16-x^2}[/spoiler]
<span class="math">\frac{1}{4}\cdot\sqrt{16-x^2}[/spoiler]
<span class="math">\frac{1}{4}\cdot\sqrt{(4-x)(4+x)}[/spoiler]

I'm really fucking bored. And tired.

[Anonymous No.3713761]

>>3713732
if your bored can you tell me how he got the 16 - x^2 out of the square root on the right side of the equation

[Anonymous No.3713774]

>>3713687
>x^2/16
>Confirmed for idiot.
Confirmed for idiot.

[Anonymous No.3713800]

>>3713761
There is a property of roots something like this (there should be something similar to this in your book):

<span class="math">\sqrt{a \cdot b} = \sqrt{a}\cdot\sqrt{b}[/spoiler]

<span class="math">\frac{1}{16}[/spoiler] was nested in the square root. If you think of <span class="math">\frac{1}{16}[/spoiler] and <span class="math">16-x^2[/spoiler] as <span class="math">a[/spoiler] and <span class="math">b[/spoiler] respectively, apply the above property and split the two apart.

When you split apart the <span class="math">\frac{1}{16}[/spoiler] term and square root it, you get <span class="math">\frac{1}{4}[/spoiler]

Don't know if that answers your question

[Anonymous No.3713810]

>>3713800
I think he may be talking about a part of the equation we can't see on the cramster website

[Anonymous No.3713851]

>>3713810
Yup, you're right. I don't want to sign up for an account right now.

All that LaTeX was pretty, though, wasn't it?

[Anonymous No.3713865]

>>3713810
let me type it for you

It is an odd answer so everyone should be able to see it, but here it is

on the right side of the equation he turns
(-1/2)[(16-x^2)^(1/2)+x(1/2)(16-x^2)^(-1/2)(-2x)]

into

(-1/2)[(16-x^2-x^2)/(16-x^2)^(1/2)]

how does he get the top part out of the square root?

[Anonymous No.3713884]

>>3713865
heres a screenshot that i circled for you guys off of the site

[Anonymous No.3713890]

>>3713884
fffffffffffffffuuuuuuuu.

forgot pic....

[Anonymous No.3713934]

aw come on you guys seriously just left before i could understand you couldnt see it? damn....

I thought you could see odd answers without an account

[Anonymous No.3714018]

>sqrt(1-x^2/16)=to (1/4)(sqrt(16-x^2)
mfw i'm 18, in year 12 and this is still extra derp maths.

confirmed for either a troll or complete moron.

[Anonymous No.3714044]

>>3714018
oh cool you can insult anonymous people

I really dont care about your opinion of me, im just kind of trying to learn maths. And if you are so great at calc, please enlighten me on how he somehow magically removed the square root. Heres a picture that shows the equation better

[Anonymous No.3714087]

nvm i fegerrd it out with ma brainz

[Anonymous No.3714182]

>>3714044
try common denominator?

[Anonymous No.3714190]

>>3714044
>>3714182
forget my post ^

He re-expressed a^1/2 as (a/a^1/2) = a^(1-1/2) get it?

[Anonymous No.3714211]

sqrt(16 - x^2) = (16 - x^2)/{sqrt(16 - x^2)}

might be confusing, but you'll need this to get common denominator

[Anonymous No.3714221]

>>3714211
^thanks for putting what I (>>3714190) said into some sense. I really need to learn LaTeX

[Anonymous No.3714230]

>>3713658
Op, check this out
sqrt(1-x^2/16)
= 4(1/4)sqrt(1-x^2/16) multiplying through by 1 = 4(1/4)
Now 4^2 = 16, so then sqrt(16) = 4
(1/4)*sqrt(16)*sqrt(1-x^2/16)
Now you know that (x^y)(z^y) = (xz)^y right? Well the squareroot is just the power half, so the same applies to the root
so we get
(1/4)sqrt(16*[1-x^2/16])
(1/4)sqrt(16-x^2)

Hope that helps.

[Anonymous No.3714238]

>>3714221

i didn't know we could use latex here...

does this works?
x = a_0 + \frac{1}{a_1 + \frac{1}{a_2 + \frac{1}{a_3 + a_4}}}
not related tho

[Anonymous No.3714310]

>>3714238
wrap it in math or eqn tags.

[Anonymous No.3714314]

>>3714190
>>3714211
>>3714044

Here