anonymous
  • anonymous
Will give medal and become fan (accepting only correct answers)! Solve for x. (x^4 - 1) / (x^3) = 0
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Please explain how you got the answer.
zpupster
  • zpupster
multply both sides by x^3 factor (x^2-1)(x^2+1) and again (x^2-1) = (x+1)(x-1) set to zero x+1=0 x-1=0 What are your roots??? (x^2+1) can never be 0 when y=0 x is?? see graph
1 Attachment
Nnesha
  • Nnesha
okay so move x^3 to the right side as ~zpupster said multiply both sides by x^3

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Okay so then it would be \[x ^{3} \times (x ^{4} - 1) \div x ^{3} = 0\]
anonymous
  • anonymous
But the x^3 can be cancelled so... the answer would be:\[x ^{4} - 1 = 0\]
anonymous
  • anonymous
Right?
Nnesha
  • Nnesha
yep you need to multiply *both sides* so 0 times x^3 = 0 now apply difference of square rule
Nnesha
  • Nnesha
\[\huge\rm a^2-b^2 =(a+b)(a-b)\]
anonymous
  • anonymous
\[(x ^{2} + 1)(x ^{2} - 1)\]
anonymous
  • anonymous
\[(x ^{2}+1)(x+1)(x-1)\]
anonymous
  • anonymous
Is that it?
Nnesha
  • Nnesha
yep right you would apply difference when there is negative sign so x^2-1 apply difference of square
Nnesha
  • Nnesha
oopps you already know
Nnesha
  • Nnesha
yes right set all 3 parentheses equal to zero solve for x
anonymous
  • anonymous
How do you find the square root of -1?
Nnesha
  • Nnesha
well do you need for this question ?
Nnesha
  • Nnesha
ohh gotcha so you will get an imaginary solution
anonymous
  • anonymous
Is it going to be \[\pm 1\]
Nnesha
  • Nnesha
\[\huge rm \sqrt{-1} = i\]
anonymous
  • anonymous
Oh like that! So the answer is x = 1, -1, and i
Nnesha
  • Nnesha
\[\sqrt{-1}=i \]
Nnesha
  • Nnesha
yep
anonymous
  • anonymous
Alright thanks again!
Nnesha
  • Nnesha
my pleasure :=)

Looking for something else?

Not the answer you are looking for? Search for more explanations.