Question

What are the real or imaginary solutions of the polynomial equation? 27x^3+125=0
I don't even know where to start and I have no idea how to do this

Answer 1

subtract, divide ....

Answer 2

what do you mean?

Answer 3

i mean that if you are traveling down i60 and hit the mcdonalds in ohio ....

what do you mean what do i mean?

Answer 4

you can eithe rsolve for x, or recall a formula for the sum of cubes

Answer 5

when I say what do you mean Im asking what d you divide and subtract

Answer 6

\[27x ^{3}+125=0\]
You want to solve for x so that when you plug this value in it will yield 0.
Can you solve for x?

Answer 7

subtract 125 and divide by 27?

Answer 8

YES!

Answer 9

that it :)

Answer 10

Try it

Answer 11

x^3=4.63

Answer 12

then are you supposed to get the cube root of 4.63

Answer 13

should be negative ... but then cbrt it for the real root

Answer 14

so x=-1.67

Answer 15

in general:\[(a^3+b^3)=(a+b)(a^2-ab+b^2)\]

Answer 16

sp the real root is

x = cbrt(-125/27) = -5/3

Answer 17

\[27x ^{3}+125=0\]
\[27x ^{3}=-125\]
\[x ^{3}=-\frac{ 125 }{ 27 }\]
\[x ^{3}=\sqrt[3]{-\frac{ 125 }{ 27 }}\]

Answer 18

amistre what number is a and what is b

Answer 19

which can be rewritten as
\[x ^{3}=\frac{ \sqrt[3]{-125} }{ \sqrt[3]{27} }\]

Answer 20

which number when you multiply it 3 times gives you -125???
Which number when you multiply 3 times gives you 27???

Answer 21

now the roots for:

9x^2 - 15x +5 should be complex and assuming i placed the 3 and 5 them correctly

Answer 22

a = cbrt(27x^3) and b= cbrt(125)

Answer 23

sorry it's suppose to be \[x=\frac{ \sqrt[3]{-125} }{ \sqrt[3]{27} }\]

Answer 24

so you get
\[x=\frac{ -5 }{ 3 }\]

Answer 25

sorry its taking me a while to reply openstudy is being really buggy for me

Answer 26

If you plug this into the original equation
\[27(-\frac{ 5 }{ 3 })^{3}+125=27(-\frac{ 125 }{ 27 })+125=-125+125=0\]

Answer 27

so it checks out

Answer 28

your solution is a real solution since you don't end up with any imaginary roots...that is we can take the cubic root of a negative number

Answer 29

yeah, its been buggy for awhile now. comes and goes lately