I need help with this question lol

[river storm]

In order to resolve the absolute you have to consider two cases:

(1): 3x + 3 ≥ 0
(2): 3x + 3 < 0

[hearty wadi]

And then

I plug in the negative numbers in the bracket?

[river storm]

You then have to solve two equations

3x + 3 = -(x² - 2x -3) if (1)
-(3x + 3) = -(x² - 2x -3) if (2)

[hearty wadi]

so:

1. x ≥ -1
2. x<-1

?

[river storm]

yea nice

[hearty wadi]

I got 1 and 0

And then what should I do?

[river storm]

for which case did you get -1 and for which 0 ?

[hearty wadi]

I simplify them
3x+3=-x²+2x+3
x=x²
I move square to other side and it becomes
√x=x
So 1=x

Is that right?

Just making sure first im on the right track

[river storm]

-x²

and then you can factor x

x = -x²
x + x² = 0
x(1 + x) = 0

and use zero product property

[hearty wadi]

So I get -(x+1)(x-3)

Im a bit confused

Can you show me the equation

[river storm]

3x + 3 = -(x² - 2x -3) if (1)
-(3x + 3) = -(x² - 2x -3) if (2)

you were almost finished with the first case

[hearty wadi]

I get this part

[river storm]

3x + 3 = -(x² - 2x -3)

-> x = -x²

-> x(1 + x) = 0

so x = 0 or x = -1

Now you have to see if those solutions are in the set: x ≥ -1

Are they?

[hearty wadi]

So you plug the negative?

[river storm]

-> x(1 + x) = 0
you mean here?

[hearty wadi]

OHHHH my bad

[hearty wadi]

I forgot to put the negative

thats why I was so confused lol

[river storm]

can answer this ?

[hearty wadi]

wait

before that

is it possible if I do this

x=-x²
I move square root to other side so it becomes
√x=-x

and then

since square root of x is still x

[river storm]

you can't do square root

[hearty wadi]

Ah

[river storm]

because -x² is negative😬

[hearty wadi]

its not possible?

[river storm]

yeah

you can factor

[hearty wadi]

Oh if its negative its not possbile?

[river storm]

this is safest option

[river storm]

yes it's actually forbidden

[hearty wadi]

Oh I never knew that

[river storm]

whaat

[hearty wadi]

Thanks for the heads up

[river storm]

ok now you know

[hearty wadi]

so you move -x² to the other side and it becomes plus

[river storm]

If you take a look at $\sqrt{x}$ you will notice that there only y $\geq$ 0 values

[wet sierra]

adonhs

[hearty wadi]

Ohhhhh

Yeah

[river storm]

yes

and then since each term has one x we can pull one

x+x²=x(x+1)

[hearty wadi]

Ok so what do you do after this?

[river storm]

you use zero product property

[hearty wadi]

what is that

[river storm]

x and (1+x) are factors and the this zero product property theorem says if x(1+x) = 0 then either x = 0 or 1+x = 0

which makes sense right

so you separetly consider now those two factors and look when they become 0

so we get x = 0 or x = -1

[hearty wadi]

Oh I see

so x is 0

and then for 1+x=0
You just move one to other side and becomes -1

I see

[river storm]

ye

now

Now you have to see if those solutions are in the set: x ≥ -1
Are they?

[hearty wadi]

Why not x<-1

why x ≥ -1

[river storm]

because we did the equation of case (1)

[hearty wadi]

Oh

[river storm]

3x + 3 = -(x² - 2x -3) if (1)
-(3x + 3) = -(x² - 2x -3) if (2)

[hearty wadi]

ok let me check

[river storm]

you said this

[hearty wadi]

so I plug in 0 for that and the answer is 3=3 which is still 3 right?

whic is 3>-1

[river storm]

yee before you plug it in

you have to make sure if they are part of your condition: x ≥ -1

because there are sometimes solutions that are not allowed

and 0 and -1 are in x ≥ -1 because 0 is greater than -1 and -1 is the same as -1

so x = 0 or x = -1 are allowed and you can plug it in to check

[hearty wadi]

Yep its the same

Is it possible if I use quadtratic formula?

[river storm]

ah yeah but it's overkill

also you have to do now case (2)

-(3x + 3) = -(x² - 2x -3) if (2) with your condition x < -1

[hearty wadi]

-3x-3=-x²+2x+3
x²-5x-6=0
(x-6)(x+1)

x=6, x=-1

Is that right?

[river storm]

correct

[hearty wadi]

then I plug it in

[river storm]

now you have to check if x = 6 and x = -1 are in x < -1

[river storm]

no

[hearty wadi]

6 is but not -1

[hearty wadi]

How do I check that?

[river storm]

x < -1 asks for all numbers that are actually smaller than -1

[hearty wadi]

So its only 6

[river storm]

-1 is not smaller than -1 but equal
6 is not smaller than -1 but greater

so there is no solution

[hearty wadi]

Oh my bad ist smaller

I thought it was bigger

my mistake

[river storm]

So x = 0 or x = -1 are the only solution because of case (1)

[hearty wadi]

My fault

[river storm]

it's important to notice these small mistakes

[hearty wadi]

Yeah my bad

I understand now

[river storm]

ok nice!

[hearty wadi]

So the answer is 0 and -1

[river storm]

yes

[hearty wadi]

Perfect

and its asking for the domain

I dont know what that is

[river storm]

well do we have any restrictions?

[hearty wadi]

Im german and it means definitionsmenge

I dont know if thats the right term in english

[river storm]

achso ja

es ist der richtige begriff

[hearty wadi]

Ach so

Dann was muss ich machen?

[river storm]

Wir haben eine Betragsfunktion rechts und links eine Parabel

Du musst nach Einschränkungen schauen

Einschränkungen haben wir wenn wir durch 0 teilen, oder die Wurzel aus negativen Zahlen ziehen

Zum Beispiel: $\frac{1}{x}$ oder $\sqrt{x}$

[wet sierra]

adonhs

[river storm]

Beim Ersten, da darf x nicht 0 sein, sonst würden wir durch 0 teilen.

Beim Zweiten darf x nicht kleiner als 0 werden, da wir sonst etwas Negatives in der Wurzel hätten.

[hearty wadi]

Wie lauten dann die Schritte und die Gleichung?

Ich hab keine ahnung haha

[river storm]

Du guckst ob eine Wurzel iwo ist oder ob durch x geteilt wird

dann wirds kritisch

aber wir haben keine Wurzel oder keine Division mit x

also ist die Domain alle reellen Zahlen

[hearty wadi]

Ach ja, und noch eine Frage

[river storm]

yes

[hearty wadi]

Vielleicht

Wenn es eine Frage gibt

[river storm]

gibts eine?

[hearty wadi]

moment

|2x-5|=1

[river storm]

was ist die frage?

[hearty wadi]

kann es -2x-5=1 sein

[river storm]

wie kommst du drauf

[hearty wadi]

Warum ist das 2x-5=-1?

nicht -2x-5=1?

[river storm]

Beide Seiten wurden mit -1 multipliziert

[hearty wadi]

Ich weiß

aber

Warum ist das nicht 2x-5=1 und-2x-5=1

[river storm]

Und nochmal: wenn du den Betrag einer Betragsfunktion auflösen willst, betrachtet du zwei Fälle:

1. Wenn der Term größer oder gleich 0 ist, dann kannst du die Striche weglassen und normal lösen (aber vergiss deine Nebenbedingung nicht)
2. Wenn der Term kleiner als 0 ist, dann machst du ein Minus um den ganzen Term mit Klammern (auch hier nicht die Nebenbedingung vergessen)

[hearty wadi]

Warum ist es 2x-5=1 und -2x-5=1

[river storm]

2x-5 = 1 wenn 2x-5 ≥ 0

-(2x-5) = -2x+5 wenn 2x-5 < 0

[hearty wadi]

ACHSO

[river storm]

Ein Minus vor der Klammer wechselt die Vorzeichen

der Terme in der Klammer

[hearty wadi]

das erklärt alles

Vielen dank

Kommst du auch aus Deutschland?

[river storm]

Ja

Deutschland kommt aber nicht aus mir

[hearty wadi]

.solved