/sci/ - Science & Math

[Anonymous No.3887968]

Hey there, would someone mind helping me out with solving this equation? thanks

[TN5 !/sci/TN5.. No.3887972]

Didn't wolfram do that right before you took your screenshot?

[Anonymous No.3887974]

square both sides then u get rid of the roots

x+8 - x+ 4 = 4
x+8 - x = 0

then factor

x+(x-8) = 0
so x = 0 or x = 8

[Anonymous No.3887986]

>>3887968

x = 8

[Lion !FvAwVR8cv6 No.3887989]

>>3887974

>no coolface.jpeg

[Anonymous No.3888000]

It did that yeah, but I wanted an explanation.
Thanks for the help /sci!

[Anonymous No.3888014]

Why do you do it, sci? Seriously, last thing we need is horde of underaged fags creating masses of threads asking for help with their bloody homework!

[TN5 !/sci/TN5.. No.3888020]

>>3888000
Fuck, dub trips! So nice I'm gonna spend 10 minute plotting a geometric proof just for you. Be right back!

[Anonymous No.3888021]

>>3887974

>x = 0

[Anonymous No.3888041]

>>3888021
ummmmmmmmm it's extraneous solution duuuuh that happens wen u hav sqare roots sometimez u no

[Anonymous No.3888045]

not under aged, just needed help thanks.

[Anonymous No.3888059]

>>3887974
>mfw most people are dumb enough to believe it.

[Anonymous No.3888056]

>>3887974
>mfw no one says anything

[Anonymous No.3888063]

>>3887974
You're an idiot.
The negative square root becomes positive thus leaving as with answer of 3

[Anonymous No.3888066]

DUB TRIPSSSS!

[Anonymous No.3888071]

>>3888059
That's good. The underaged believed him, so he will get 0!

[TN5 !/sci/TN5.. No.3888348]

>>3888020
Geometric proof incoming.

We can look for <span class="math">\sqrt{8}[/spoiler] instead of <span class="math">x[/spoiler] since <span class="math">x\geq 4[/spoiler] from the equation.

Thus, we start by drawing the red vertical line segment of length 2 that we call <span class="math">[AB][/spoiler]. We draw 2 red circles of radius 2 centered in <span class="math">A[/spoiler] and <span class="math">B[/spoiler]. We draw the perpendicular to <span class="math">[AB][/spoiler] in <span class="math">A[/spoiler] and call <span class="math">C[/spoiler] its left intersection with the circle centered in <span class="math">A[/spoiler].
Pythagoras gives <span class="math">BC=2\sqrt{2}[/spoiler], and we report that distance on the line <span class="math">AB[/spoiler] to obtain point <span class="math">D[/spoiler] such that <span class="math">BD=2\sqrt{2}[/spoiler] and <span class="math">A[/spoiler] is between <span class="math">B[/spoiler] and <span class="math">D[/spoiler].
Now we draw the line segment <span class="math">[DF][/spoiler] perpendicular to <span class="math">[BD][/spoiler] and call <span class="math">\sqrt{x}[/spoiler] its length.
Pythagoras gives <span class="math">BF=\sqrt{x+8}[/spoiler].
We call <span class="math">E[/spoiler] the intersection of <span class="math">[BF][/spoiler] with the circle of radius <span class="math">2[/spoiler] centered on <span class="math">B[/spoiler].

[TN5 !/sci/TN5.. No.3888355]

>>3888348
<span class="math">BE=2[/spoiler], thus as <span class="math">\sqrt{x+8}=BF=BE+EF=2+EF[/spoiler], we have that <span class="math">EF=\sqrt{x-4}[/spoiler] for the equation to be satisfied.
Now we draw the circle of radius 2 centered on <span class="math">E[/spoiler], and the perpendicular to <span class="math">[BF][/spoiler] in <span class="math">E[/spoiler]. The intersection between these two forms, with <span class="math">E[/spoiler] and <span class="math">F[/spoiler], a triangle of sides <span class="math">2[/spoiler], <span class="math">\sqrt{x-4}[/spoiler], and, by Pythagoras, <span class="math">2\sqrt{2}[/spoiler]. This intersection must therefore be on the circle of radius <span class="math">2\sqrt{2}[/spoiler] centered on <span class="math">F[/spoiler].
The first picture shows an attempt with a wrong value of <span class="math">\sqrt{x}[/spoiler].
The second picture shows a working attempt where we can see that the intersection is in <span class="math">D[/spoiler] and that therefore, <span class="math">\sqrt{x-4}=2[/spoiler] so <span class="math">x=8[/spoiler]. Checking that <span class="math">x=8[/spoiler] is a solution of the equation is then trivial. Uniqueness follows from the fact that <span class="math">\sqrt{.+8}-\sqrt{.-4}[/spoiler] is strictly decreasing for on <span class="math">(4;\infty)[/spoiler].

[TN5 !/sci/TN5.. No.3888439]

>>3888355
C'mon guys, it's a geometric resolution of an algebraic equation with squares roots and all! Geometric proofs are awesome! They are almost as awesome as fixpoint theorems or topological views of linear algebra!

Give me some lovin'!

[Anonymous No.3888527]

>>3888439
WOW ITS NOT AS SIMPLE AS >>3887974

[TN5 !/sci/TN5.. No.3888700]

>>3888527
It's not as troll either. Not saying it's not troll at all, but the maths are (I think) sound. And actually, if I were the teacher and the student gave me this geometric proof, I'd WTF so much I'd give him a good mark for sure. Like... WHY U DID TAHT?!

[Anonymous No.3890572]

>>3887974
FUCK NOPE.

square the whole equation and you get:

x + 8 + x - 4 =4
2x + 4 = 4
2x = 0
x=0

[TN5 !/sci/TN5.. No.3890807]

>>3890572
Square the whole equation and you get:
<span class="math">(\sqrt{x+8}-\sqrt{x-4})^2=4[/spoiler]
(just so you know, <span class="math">(a-b)^2[/spoiler]=a^2+b^2-2ab)

Then
<span class="math">(x+8)+(x-4)-2\sqrt{x+8}\sqrt{x-4}=4[/spoiler]
<span class="math">2x+4-2\sqrt{(x+8)(x-4)}=4[/spoiler]
<span class="math">x=\sqrt{(x+8)(x-4)}[/spoiler]
<span class="math">x^2=(x+8)(x-4)[/spoiler]
<span class="math">x^2=x^2+4x-32[/spoiler]
<span class="math">4x=32[/spoiler]
<span class="math">x=8[/spoiler]
(where the fact that <span class="math">x\geq 4>0[/spoiler] is used several times to preserve equivalence)

I'm not sure why you necro bumped this thread to say something stupid, but if it was trolling, it certainly deserves a 3/10 for the effort.