Feel like a moron

Feel like a moron
can anybody solve and explain?
Been a while since I learned more complex math and could really use some freshening up

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I haven't been in school in years, but wouldn't that be x+3?

solving for nothing KEK

there's nothing to solve faggot

what do you want to solve it's not even an equation
well if you want i can tell you x = 148974,3

nope, you have to solve the sum before the root

Nope. If you take x+3 and square it, you get (x+3)(x+3) which equals x^2+6x+9
This expression is already in basically simplest form.

sorry, solve for x

Nope. Power is not distributive across addition. This would be some square root of i shit if there was anything to solve...

solve for zero you idiots

No. Since (x+3)² would be x² + 6x + 9

Go and use wolframalpha.com Sup Forumsro

F(a)=g^2○t

Not 100% but find difference between two squares so
(X+3)(X-3)^1/2

Then wouldn't your solutions be root 3 and you can't have a negative root 3 (no imaginary numbers for this)

you forgot 2 more parenthesis stupidhead

Yea I know you nigger

sqrt(x^2+9)=0 《》x€(3 i, -3i)

And, dear OP: GTFO!

In that case
x=3i

√(x^2 + 9) = 0

Square both sides

x^2 + 9 = 0

x^2 = -9

x = √-9

x = 3i

It's not a difference of squares

There is literally nothing to solve. It's just a mathematical expression. There's no equation, there is nothing to extrapolate.
You can place any x you want, and it'll give you a number. That's like saying "7 is a number".

You are right but I think we were just supposed to assume it's equal to 0. OP is an idiot either way

> OP is an idiot either way

Surely. People with working brains uses wolframalpha.com

x ε IR

Or x is any reall number.

/Thread

x=(-3)

who said it equals zero?
What you're finding is the roots
And you only found one of them
x=3i V x=-3i

Excalibur
Excalibur
From the United King
I'm looking for him
I'm going to California
Excalibur
Excalibur
From the United King
I'm looking for him
I'm going to California
Excalibur
Excalibur
Excalibur

Go back to primary school

Cant solve this as it isnt an equation.
You can simplify it though.

Its the same as x-3

Shouldn't it be (x+3)*(x-3) to get that small of a polynomial

(x+3)(x-3) = x^2-9

People with working brains don't need wolfram alpha to do basic math.

yeah it's been a decade since i was in school, I'm still closer than most these people...
But we still don't know what we're supposed to solve for so whatever.

No one said it equals 0, I just assumed cause op is retarded. And yes I did leave out the negative answer. My mistake

3x thanks

here
x=0 the answer is +/-3
x=sq.root 7, the answer is +/-4
my personal favorite, x=4, the answer is +/-5

well excuse me for not getting it as fast as you, since WA only gave the solution, but not the basic steps and explanation for people not as good in math as you
sorry for asking a question

Before you start to cry, notice, that i^2=-1 or i=sqrt(-1) and it's imagine part of complex number

It has no real roots though
>(x^2+9)^1/2=0
>x^2+9=0^2(=0)
>A=1 B=0 C=9
>(B^2-4AC)^1/2=6i
>(B±6i)/2=(B/2)±3i=ROOTS

Thats
ROOTS: ±3i
>just to simplify for those retards who couldnt substitute the 0 in

Kingdom*
Kingdom*
you for a dom-dom part
and remember he says it Excaleebur lol

>It has no real roots though
So what.

I didn't say it had real roots? I was the one who did to step by step showing the root to equal 3i. I just forgot the -3i as well.

You usually solve for real roots, at least that's the default assumption.

It's okay to not understand the steps but just make sure you write out the entire problem, cause the way you did it just doesn't give much. We have to know what that expression is equal to in order to do anything

alright, I'll keep that in mind, thanks

It depends on the topic. In algebra 2 classes there's a section devoted to imaginary numbers so when doing that section you will find the imaginary roots, because that's the purpose.

>You usually solve for real roots, at least that's the default assumption.
No it's not; in high school algebra 1 it is but there isn't really some 'default' in math. The fundamental theory of algebra puts no preference on what roots are 'important' in ℂ.

theorem*

THE STEPS:
[1] Square both sides. (x^2+9=0)
[2] Substitute b & c into quadratic discriminant (b^-4ac)
[3] the discriminant is negative so there are no real roots (-36)
[4] make discriminant positive to find complex roots (36)
[5] substitute discriminant into quadratic formula ((-b±(discriminant)^1/2)/2a) so ±6/2
[6] get roots (remember to add i, we turned the discriminant earlier so these are not real) =±3i

You can skip the discriminant and just plug everything into the quadratic formula.

you get the discriminant to check if your result will be real or imaginary, it's not an extra step since you have to calculate it in the formula anyway.

"solve 2"
>other people are idiots

There's no point in doing all of that. Just do some basic algebraic manipulation and you'll get the answers. Square both sides, move the 9 over, and take the square root. Then account for the imaginary aspect.

You fucking idiot

Yes, the discriminant is in the radical but there is no real reason to 'check' because checking if ℑm[Δ]≠0 doesn't do anything; all it does it make it look like there is some stigma against complex values. Just do the arithmetic and pull i out of the root when you have to.

True, I'm maths noob.
However you cannot always do that so it's better to know a universal method that'll work in all cases.
Just one thing:
so you get to x^2=-9
root both sides to get x=3i
where does -3i come from? Surely -3i^2=9 so it's not a root? Im making an idiot mistake somewhere.

Meant to write ℑm[(Δ)^(1/2)]≠0

was thinking the same at first, -3 and 3. But it's x^2 + 9, not x^2 - 9. You can't solve this for zero. Unless we add imaginary.

(x - 3i)*(x + 3i) = x^2 - x3i + x3i - (i^2)*9
= x^2 + 9

The statement is therefore zero when x = +-3i

>However you cannot always do that so it's better to know a universal method that'll work in all cases
The quadratic formula is that method. Plugging a,b,c directly into it will give you both roots, up to multiplicity, of any polynomial of degree 2 or less.

The -3i comes from taking the square root. When you take a square root of something you are always technically getting 2 answers, positive and negative. Basically if you take the square root of 4 you get 2, because 2^2 is 4. But -2^2 (-2 * -2) is also 4. So you have 2 solutions

This

Just keep practicing OP. That's all it takes with math, patience and practice. Look up patrickjmt on YouTube for help. You'll get it

ok so Im fucking up really bad at some point, teach me master.
(-3sqrt(-1))*(-3squrt(-1))=8-6sqrt(-1)

Okay what you typed makes no sense. You need a run down on how complex numbers work. Gimme a few minutes.

I don't really understand why you are doing that or how you got that answer from doing that

-3^2=9
sqrt(-1)^2=-1
-3*sqrt(-1) = -3i
-3*sqrt(-1) = -3i
9-1-3i-3i = 8-6i

Okay...but why? 3i and -3i are your roots. That's it. I'm not sure why you are subtracting it from 9 and 1...

http://www.wolframalpha.com/input/?i=%E2%88%9A%28x^2%2B9%29

This is not about OPs problem.
The point is if -3i is a root of -9 then -3i squared should be -9, no? But I'm just getting 8-6i out of it...

You aren't doing anything. 3i and -3i are the answers.

That product is super wrong.

Just read this:purplemath.com/modules/complex.htm

>root of -9
Numbers don't have roots, they're numbers.

(-3i)^2

i^2 = -1

-3^2 = 9

9*-1 = -9

>Numbers don't have roots, they're numbers.
Never mind, I was thinking of functional roots.

Just took a test on this shit. You guys are morons

√x²+9
√x²+9 = 0
- set equal to zero
(√x²+9)² = 0²
- square to get rid of radical. Pretend the radical encompasses +9 also
X² + 9 = 0
X² = 9
- subtract 9
√x² = √9
- Radical on both sides to get ride of square
x = ±3
- X is equal to plus/minus 3

...

Should be plus/minus 3i because subtract nine. My bad

did you just...

Yes we are morons because we figured it out like 20 minutes ago. Also you are wrong anyway

complex number equation in most basic form

How's summer so far?

Grade 12 student that just went over this for review.

You square root each number seperately.
X+3. You can't get a simpler answer

...

Kek

I failed for your bait hahahaha

Mathematician here.


Set sqrt(x^2+9) equal to zero
>sqrt(x^2+9) = 0

Square both sides canceling out the square root
>now you're left with x^2+9=0

Subtract nine from each side
>x^2=-9

Take the square root of both sides to undo x squared
>now you're left with x= sqrt(-9)

>But wait user, you can't take the square root of a negative number!

Fuck you nigger, we can solve this shit using IMAGINARY goddamn numbers

>Therefore, x=3i with a multiplicity of two since you had to square the zero in a previous operation

You're a shitty mathematician.

>x=3i with a multiplicity of two
Wrong. The roots are 3i and -3i, both with multiplicity one.

And you should already be able to tell that the other root is the complex conjugate of the one you found because all the polynomial coefficients are real.

You nine gagging motherfucker

Just simplify it as much ad possible? Because you can't solve for X.

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Almost 100% of guys here are idiots... there is no equation

OP later specified that he was hunting for roots of the expression.

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