A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
Find the value of x: 2x^3  5x^2  x + 20 = 0
anonymous
 4 years ago
Find the value of x: 2x^3  5x^2  x + 20 = 0

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0would you like to go for a trignometric solution ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I just need the value of x :o the answers have 2 x values. Its a whole different equation but Ive added everything and I just needed help on finding the x values of this because It confused me

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if you want soln,,i can provide hint : see shubhamsrg's soln : http://openstudy.com/users/shubhamsrg#/updates/4fed5f8ae4b0bbec5cfcc1ac

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0first you have to convert the eqn into depressed but..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.01: x= 4/3, x= 5 2: x= 4/3, x= 5 3:x= 4/3, x= 5 4:x= 4/3, x= 5

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=2x%5E3%2B5x%5E2x%2B20%3D0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I tried plugging it into wolframalpha already.. didnt work _

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0didn't work! there the ans. given is 1.6

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok that gives me one answer but :o

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I guess ill just try Plugging 5 and negative 5 into the original thing.. Thanks :p

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Wait can Someone actually Help me? Cause Ive tried plugging the two 5's in but Its not equaling out. @Wired @JohnHanShanghai , the original equation is this,

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342092140095:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0have you tried 2 and minus 2?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0But thats not a possible answer?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think Im retarded because either 5 or 5 have to work . But I tried and of course I did something wrong

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Wolfram says 4/3 and 5. Must've done something wrong when solving the equation.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0there are 2 answers with 4/3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Wait where does it say 5?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=%282%2F%28x2%29%29%2B%287%2F%28x%5E24%29%29%3D5%2Fx

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well I didnt put int he original equation but thank you ! Again lol (:

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The correct eq. is 3x^211x20=0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\frac{2}{x2}+\frac{7}{x^{2}4}=\frac{5}{x}\] \[\frac{x+2}{x+2}*\frac{2}{x2}+\frac{7}{x^{2}4}=\frac{5}{x}\] \[\frac{2x+4+7}{x^{2}4}=\frac{5}{x}\] \[2x^{2}+11x = 5x^{2}20\] \[3x^{2}11x20 = 0\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0When in doubt, go back to the beginning :)
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.