A community for students.
Here's the question you clicked on:
 0 viewing
farmergirl411
 2 years ago
Solve for x
x^2(x3)^24(x3)^2=0
farmergirl411
 2 years ago
Solve for x x^2(x3)^24(x3)^2=0

This Question is Closed

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.0Can you rearrange that so it is a product? If you can, then set each component of the product = 0 and solve for x.

farmergirl411
 2 years ago
Best ResponseYou've already chosen the best response.0um not totally sure

farmergirl411
 2 years ago
Best ResponseYou've already chosen the best response.0can you solve it another way I am totally lost with this problem

farmergirl411
 2 years ago
Best ResponseYou've already chosen the best response.0can you help please

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.0What do you get if you multiply \[(x^24)(x3)\]?

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.0Let's make a substitution to make it easier to see what is going on. Instead of \[x^2(x3)^2  4(x3)^2 = 0\]We'll let u = (x3) and write\[x^2u^24u^2 = 0\]Can you factor that at all?

ambius
 2 years ago
Best ResponseYou've already chosen the best response.0We know that: \[= (\sqrt{x^2}x \sqrt{x^2}3)^2  (\sqrt{4}x3 \sqrt{4})^2\]

ambius
 2 years ago
Best ResponseYou've already chosen the best response.0no substitutions necessary. : )

ambius
 2 years ago
Best ResponseYou've already chosen the best response.0\[0 = (x^2  3x)^2  (2x  6)^2\]

ambius
 2 years ago
Best ResponseYou've already chosen the best response.0\[(x^23x)^2 = (2x6)^2\] then root both sides, simplify, and solve for x.

valentinekid
 2 years ago
Best ResponseYou've already chosen the best response.0@whpalmer4 had the right idea. Factor (x3)^2 out: \[(x^24)(x3)^2=0\]

whpalmer4
 2 years ago
Best ResponseYou've already chosen the best response.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 \[u^2(x^24)=0\]and then when you plug \(u = (x3)\) back in, you've got \[(x3)^2(x^24) = 0\]Yes, someone more experienced at factoring of this sort will notice directly that \[x^2(x3)^2  4(x3)^2 = 0\]factors to \[(x3)^2(x^24) =0\]
Ask your own question
Sign UpFind more explanations on OpenStudy
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
 Engagement 19 Mad Hatter
 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.