Quantcast

A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

farmergirl411

  • 2 years ago

Solve for x x^2(x-3)^2-4(x-3)^2=0

  • This Question is Closed
  1. whpalmer4
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Can you rearrange that so it is a product? If you can, then set each component of the product = 0 and solve for x.

  2. farmergirl411
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    um not totally sure

  3. farmergirl411
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    can you solve it another way I am totally lost with this problem

  4. farmergirl411
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    can you help please

  5. whpalmer4
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    What do you get if you multiply \[(x^2-4)(x-3)\]?

  6. whpalmer4
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Let's make a substitution to make it easier to see what is going on. Instead of \[x^2(x-3)^2 - 4(x-3)^2 = 0\]We'll let u = (x-3) and write\[x^2u^2-4u^2 = 0\]Can you factor that at all?

  7. ambius
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    We know that: \[= (\sqrt{x^2}x- \sqrt{x^2}3)^2 - (\sqrt{4}x-3 \sqrt{4})^2\]

  8. ambius
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no substitutions necessary. : )

  9. ambius
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[0 = (x^2 - 3x)^2 - (2x - 6)^2\]

  10. ambius
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[(x^2-3x)^2 = (2x-6)^2\] then root both sides, simplify, and solve for x.

  11. valentinekid
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @whpalmer4 had the right idea. Factor (x-3)^2 out: \[(x^2-4)(x-3)^2=0\]

  12. whpalmer4
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @ambius no substitution was *necessary*, but people often have trouble spotting that they can factor out a binomial rather than a single term, until they've done it a few times. Spotting that one can factor out a single term is something that many do more easily. It's easy to see that \[x^2u^2 - 4u^2 =0\]can be factored to \[u^2(x^2-4)=0\]and then when you plug \(u = (x-3)\) back in, you've got \[(x-3)^2(x^2-4) = 0\]Yes, someone more experienced at factoring of this sort will notice directly that \[x^2(x-3)^2 - 4(x-3)^2 = 0\]factors to \[(x-3)^2(x^2-4) =0\]

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

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.