Im stupid

[austere sleet]

sq root x+7=x-5

[near dawn]

Is it sqrt (x), or sqrt (x+7)

[austere sleet]

the x+7under the radical

[near dawn]

Start by squaring both sides of the equation, so all terms are real and rational

[austere sleet]

x^2-10x+25

=x+7

2^2-11x+18

x^2

[idle cape]

you should try to square root(x+7) and (x-5)

to take out all roots so its easier to work with

[austere sleet]

i got 9,2

[steady knoll]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[austere sleet]

did i get it wrong

[idle cape]

im gonna check

youre solutions are 9 and 2

or 9,2

[austere sleet]

Im gonna kill myself for real

the answers are programmed wrong

😩

[idle cape]

u got it right

its 9 and 2

[austere sleet]

It says ten for multiple questions

i doubted myself whole time

[idle cape]

its 9 and 2

[austere sleet]

im cooked

[idle cape]

trust me

explain to me how u got your answers

[fleet dirge]

Your only solution is 9

[idle cape]

technically ye

[fleet dirge]

Well no not technically

Square root can't be negative

[idle cape]

if they work with imaginary numbers

[fleet dirge]

This is actually still not true

There's nothing you can do to get a square root to give a negative real number

Its only when you invert x^2

[idle cape]

right

sorry its getting late

[fleet dirge]

It's confusing because you would think that the square root is a perfect inverse of x^2

[fleet dirge]

Don't worry it's easy to forget

Just making sure we don't have any incorrect solutions because we forgot

[austere sleet]

I hate my life

😩

[idle cape]

youre saying that sqrt of 9 cant equal -3

if we take 2 as an answe

[fleet dirge]

Don't worry math will get much worse

[fleet dirge]

Yes

[idle cape]

hmm thank you

[fleet dirge]

$\sqrt{9} \neq -3$

[wide halo]

hiidostuff

[austere sleet]

I know

[fleet dirge]

But (-3)^2 = 9

[austere sleet]

I trying to learn advanced stuff so i can rob banks

I gotta crack aes 256 encryption

[fleet dirge]

So real

[austere sleet]

The probability matrix on a sliding cipher tho must be insane

[fleet dirge]

I'm learning advanced stuff just bc I'm a nerd but

[idle cape]

juts learn how to derivates e^x and youre all set

[fleet dirge]

A large part of complex numbers comes from this fact

Pretty sure you can necessitate the definition of holomorphism by giving an example of a function that requires a more rigorous analysis on its continuity

And it involves that

[pale dust]

I gonna die

[shadow musk]

If you start from an equation like $\sqrt{x+7} = x-5$,

you can recognise that the solutions must be restricted,

since we are solving over $x\in\mathbb{R}$, to values satisfying

two requirements: firstly, $x + 7 \ge 0$ (as no negative

real has a real square root); secondly, $x - 5 \ge 0$ (as

no real has a square root that is negative.

In other words, the solutions over this equation over

the reals must be both $x \ge -7$ and $x \ge 5$, which

combine to form a domain from which all the solutions

must come: $x\in\big[5,,\infty\big)$.

Rearranging the equation cannot cause solutions

outside this domain to become possible, and so any

such apparent solutions must be rejected.

[wide halo]

Jay

[austere sleet]

WHAT

WHAT KINDA MAGIC IS THIS

IM COOKED

[fleet dirge]

It's all just domain and range

You gotta know what types of a numbers a function can eat and spit out

[shadow musk]

\begin{align*} \sqrt{x+7} &= x-5 \qquad \text{noting $x\ge 5$} \ \left(\sqrt{x+7}\right)^2 &= \left(x-5\right)^2 \qquad \text{on squaring} \ x + 7 &= x^2 - 10x + 25 \qquad \text{recalling $(A-B)^2 = A^2 - 2AB + B^2$} \ 0 &= x^2 - 11x + 18 \ 0 &= (x-2)(x-9) \qquad \text{on factorising} \ x - 2 = 0 \qquad &\text{or} \qquad x - 9 = 0 \ x = 2 \qquad &\text{or} \qquad x = 9 \ \text{Hence,} \quad x &= 9 \qquad \text{is the only solution that is valid} \ &\qquad \qquad \text{as only it satisfies the requirement that $x\ge 5$.} \end{align*}