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Mathematics 23 Online
OpenStudy (anonymous):

how to solve x cubed plus 27 equals 0

OpenStudy (jhannybean):

\[x^3 +27 = 0\]subtract -27 from both sides of the equation first. Take the cube root of both sides of the function. You will find your answer.

OpenStudy (anonymous):

\[x^2+27=0\] one answer you get by thinking the other two by factoring the sum of two cubes and using the quadratic formula for the second factor

OpenStudy (anonymous):

the answer is 3 then

OpenStudy (dtan5457):

the cube root of -27 is -3. -3 times -3=9 times -3 again is -27.

OpenStudy (jhannybean):

Not just 3. \(3 \cdot 3 \cdot 3 \ne -27\)

OpenStudy (jhannybean):

@dtan5457 no answers.

OpenStudy (anonymous):

oh no solution

OpenStudy (jhannybean):

False.

OpenStudy (anonymous):

oh no

OpenStudy (anonymous):

you want a number that when cubed give you \[\huge \color{red}-27\]

OpenStudy (anonymous):

oh gotcha i got tht one but thrs more lol

OpenStudy (dtan5457):

^-1

OpenStudy (jhannybean):

Whereas multiplying it an even number of times will result in a positive answer: (-1)(-1)(-1)(-1)

OpenStudy (jhannybean):

Any time you multiply a negative an odd amount of times, you will end up with a negative answer, i.e (-1)(-1)(-1) =-1

OpenStudy (anonymous):

find the real solutions of x to the fourth plus 2x squared equals 5 using a graphing utility. round each solution to the nearest hundredth.

OpenStudy (jhannybean):

Can you write out your function first?

OpenStudy (anonymous):

\[x ^{4} + 2x ^{3} = 5\]

OpenStudy (jhannybean):

Correction: \(x^4 +2x^2=5\)

OpenStudy (anonymous):

on my paper its 2x cubed

OpenStudy (jhannybean):

davidellis17 find the real solutions of x to the fourth plus 2x squared equals 5 using a graphing utility. round each solution to the nearest hundredth. 2 minutes ago On here you hae written 2x squared, so that is why I corrected you.

OpenStudy (jhannybean):

But ok, \(x^4 +2x^3=5\)

OpenStudy (anonymous):

oh srry bout tht

OpenStudy (jhannybean):

Pull out the LCM, between \(x^4\) and \(x^3\) which one would it be?

OpenStudy (anonymous):

3

OpenStudy (jhannybean):

Hmm... I'm not sure this function has any real solutions.

OpenStudy (jhannybean):

@satellite73 ? A little help here.

OpenStudy (anonymous):

it would be so much easier if i had someone here lol

OpenStudy (anonymous):

what about \[20 -\frac{ x }{ c } = d\] solve for x

OpenStudy (fibonaccichick666):

so do you have a calculator?

OpenStudy (anonymous):

yes

OpenStudy (fibonaccichick666):

what type?

OpenStudy (anonymous):

ti-30xa

OpenStudy (fibonaccichick666):

that's not a graphing calculator... darn

OpenStudy (fibonaccichick666):

ok so you have to do it the algebraic way instead of using a utility

OpenStudy (anonymous):

alright

OpenStudy (fibonaccichick666):

\[ x^4+2x^3=5\] SO you have that, now what is the biggest number/variable expression you can pull out of BOTH \(x^4\) and \(2x^3\)?

OpenStudy (anonymous):

oh tht was a different prob but i guess x squared

OpenStudy (anonymous):

x cubed

OpenStudy (fibonaccichick666):

which is it xsquared or x cubed?

OpenStudy (anonymous):

x cubed lol

OpenStudy (fibonaccichick666):

okee dokey so do you know what that looks like when you pull it out?

OpenStudy (fibonaccichick666):

can you "un"factor x^3?

OpenStudy (anonymous):

well prob x plus 2x = 5

OpenStudy (fibonaccichick666):

not quite,

OpenStudy (anonymous):

well im stuck

OpenStudy (fibonaccichick666):

it will be \[x^3(x+2)=5\] You cannot forget to write the whole equation, and what is x^3/x^3?

OpenStudy (anonymous):

graph \[y = \sqrt{x+5} -3\] find the domain and range

OpenStudy (fibonaccichick666):

ya know what, let's just do it how the problem asked us to. Here use this as your graphing utility. https://www.desmos.com/calculator

OpenStudy (anonymous):

ohh i have tht on my cromebook

OpenStudy (fibonaccichick666):

just graph your eq and find the zeroes, it's way easier

OpenStudy (fibonaccichick666):

if you had x^2 as you said prior it would be easy, but unfortunately, we do not

OpenStudy (anonymous):

this stuff is not funn

OpenStudy (fibonaccichick666):

to someone who doesn't get to do easy stuff often it is

OpenStudy (anonymous):

lol thats rights

OpenStudy (anonymous):

i just wish someone was here helping me

OpenStudy (fibonaccichick666):

well, that's what your actual teacher is for...? Or you could always hire a tutor

OpenStudy (anonymous):

ok i hire you to my house noww lol jk

OpenStudy (fibonaccichick666):

you couldn't afford me :P

OpenStudy (anonymous):

funny but prob true lol

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