/sci/ - Science & Math

[Anonymous No.2743860]

Are you?

[Anonymous No.2743868]

Rationalization. That's a tough one.

[Anonymous No.2743871]

x + 1 = 0
x - 2 = 0
4x - 11 = 0

3 equations, 1 variable
thus unsolvable

Q.E.D.

[Implying No.2743873]

Yes.

[Anonymous No.2743875]

x = 3

[Anonymous No.2743876]

>>2743871
0/10

[Anonymous No.2743880]

>>2743875
How did you figure that out?

[Anonymous No.2743884]

>>2743860
but what if i squared both sides then solved for x?

[Anonymous No.2743888]

>>2743880

guess and check like a boss

[Anonymous No.2743894]

>>2743888
Damn
I wanted to know how to solve algebraically :/

[Anonymous No.2743897]

Squaring is the dumb highschooler way of solving, there has to be a different method. Ask yourself, how would have Gauss addressed this problem?

inb4 he would have squared

[Anonymous No.2743902]

>>2743897
Squaring makes no sense, it's too complicated

[MusicPiggy !!iMBB9yBnJYG No.2743918]

>>2743888
Seriously. This is all you need to solve that.

[Anonymous No.2743928]

Numerical methods here, symbolic algebra is a fagot.

[Anonymous No.2743934]

0 squared is still zero, just square both sides
(x+1)-(x-2)-(4x-11)=0
Imma go with >>2743871

[Anonymous No.2743946]

x=23/7

you need to square.

[Anonymous No.2743958]

Everyone in this thread who squared anything
>areyouatrollorjustdumb.jpg

[Anonymous No.2743961]

>>2743897

Here

I think I got this.

instead of equating to 0 make it equal to f(x)

The function is surely defined for x >11/4 which is 2.75, we can apply Bolzanos theorem from 2.75 till positive infinity, when x=11/4 f(x) is positive, when x=4 f(x) is -sqrt2 which means there has to be a 0 between 2.75 and 4, then you plug 3 and see it gives 0

coolface.tga

[Anonymous No.2743973]

HS faggot here.

Why is it that you can't modify the equation using multiplication/division when one side is zero? I mean, is there an algebraic reason, or is it just an exception to the rule?

[Anonymous No.2743986]

problem?

[Anonymous No.2744022]

CALCULUS

[Anonymous No.2744025]

>>2743934
urnotdoinitrite

let me lend a hand?

(x+1)^.5 - (x-2)^.5 = (4x - 11)^.5

square to get

x+1 + 2(x+1)^.5(x-2)^.5 + x-2 = 4x-11
(x^2 - x - 2)^.5 = x - 5
x^2 - x - 2 = x^2 - 10x + 25
9x = 27
x=3

also, >>2743875


>lrn2algebra

[Anonymous No.2744027]

(sqrt(x+1) - sqrt(x-2) - sqrt(4x - 11))(sqrt(x+1) - sqrt(x-2) + sqrt(4x - 11)) = 0


(sqrt(x+1) - sqrt(x-2))^2 - (4x - 11) = 0

x + 1 - 2 sqrt(x+1)sqrt(x-2) + x - 2 - (4x - 11) = 0

2x - 10 + 2 sqrt(x+1)sqrt(x-2) = 0

(2x - 8 + 2 sqrt(x+1)sqrt(x-2))(2x - 8 - 2 sqrt(x+1)sqrt(x-2)) = 0

(2x-10)^2 - (x+1)(x-2) = 0

3x - 39x + 102 = 0

prolly made some errors but the method is good

[Anonymous No.2744034]

I hope you guys know squaring it gives you the wrong answer...

[MusicPiggy !!iMBB9yBnJYG No.2744050]

>>2744025
OR we know x must be a relatively low number because x+1 and x-2 are both perfect squares. See that 3+1=4, 3-2=1. Done.

[Anonymous No.2744059]

>>2744034
only gives you wrong answer if urnotdoinitrite

(3+1)^.5 - (3-2)^.5 - (12-11)^.5 = 0
2 - 1 - 1 = 0

Gameset

[Anonymous No.2744058]

2 x+sqrt(x-2) sqrt(4 x-11) = 7
9 x-27 = 0
x=27/9
how the fuck are you suppose to divide 27 by 9?

unsolvable

[Anonymous No.2744066]

>>2744058
haha

[Anonymous No.2744085]

So here's an actual proof. We can find that <span class="math">x=3[/spoiler] is a solution. We prove that there are no others where <span class="math">x\in\mathhb{R}[/spoiler]. Rearranging the equation we get <div class="math">\sqrt{4x-11} = \sqrt{x+1}-\sqrt{x-2}.</div>
Define <div class="math">K=\sqrt{x+1}-\sqrt{x-2}</div>so that
<div class="math">3=K\sqrt{4x-11}.</div>

Suppose that <span class="math">x<3[/spoiler], then <span class="math">K<3[/spoiler] (since sqrt is monotonically increasing) and <span class="math">\sqrt{4x-11}<1[/spoiler], which contradicts <span class="math">3=K\sqrt{4x-11}.[/spoiler]

Suppose that <span class="math">x>3[/spoiler], then <span class="math">K>3[/spoiler] (since sqrt is monotonically increasing) and <span class="math">\sqrt{4x-11}>1[/spoiler], which contradicts <span class="math">3=K\sqrt{4x-11}.[/spoiler]

Have a good day y'all

[Anonymous No.2744087]

>>2744085
meant to say "We prove that there are no others where <span class="math">x\in\mathbb{R}[/spoiler]"

[Anonymous No.2744092]

>>2744085
correction: "define <div class="math">K=\sqrt{x+1}+\sqrt{x-2}</div>"

[Anonymous No.2744093]

>>2744085
Unrelated question but,

How do you use latex?

[Anonymous No.2744105]

surround your code with \[math\] \[\/math\] tags

[Anonymous No.2744113]

(x + 1) - (x - 2) - (4x - 11) = 0
x + 1 - x + 2 - 4x + 11 = 0
-4x + 14 = 0
x = 3,5

[Anonymous No.2744141]

>>2744113
5 is not a solution

[Anonymous No.2744168]

>>2744113
this is what I did.. but its not 2 answers 3, and 5. its 3.5. kind of confusing there

[Anonymous No.2744190]

>>2744141
he made a mistake an thought the answer was 3.5 (three point five)

[Anonymous No.2744312]

Why does squaring give you the wrong answer?

[Anonymous No.2744337]

>>2744312
Same reason that
(a+b)^2 =/= a^2 + b^2

[Anonymous No.2744345]

>>2744337

How is it the same?

[Anonymous No.2744355]

he means <div class="math">(a+b)^2 \ne a^2 + b^2</div>

[Anonymous No.2744364]

>>2744345
You have several terms on the left side, you can't simply use the argument of "I'll just square everything and remove the radicals". It's the same thing as trying to square (a+b) and getting a^2 and b^2 as your terms.

It doesn't work that way.

[Anonymous No.2744414]

Math noob here. Am I correct when saying that:

(a+b)^2 = a^2 + b^2 + 2ab

[Anonymous No.2744456]

Bring one of the roots over the other other side, square, solve for x. This is shit you would see in grade 9.

[Anonymous No.2744487]

Testing latex...

\[math\]x=(-b+sqrt(b^2-4*a*c)/2*1\[\/math\]

[Anonymous No.2744512]

\[math\]
x^2+y^2=exp(2)
\[\/math\]

[Anonymous No.2744519]

>>2744414
Yes, that's correct.

[Anonymous No.2744529]

>>2744487
>>2744512


errrr, guize? how do i latex?

[Anonymous No.2744541]

>>2744529
Drop the backslashes in front of the square brackets.

[Anonymous No.2744559]

it's like UBcode

[.math][./math]

but without the dot.

[Anonymous No.2744568]

<span class="math">x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}
[/spoiler]

Captcha: abloper example

Let's hope not

[Anonymous No.2744576]

+1 internets for whoever can tell me what this is used for...

<span class="math">F=\frac{\pi ^2EI}{K^2L^2}[/spoiler]