Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
sauravshakya
Group Title
Another Fun Question
Find all the real solutions:
x^14=7+x^2
 one year ago
 one year ago
sauravshakya Group Title
Another Fun Question Find all the real solutions: x^14=7+x^2
 one year ago
 one year ago

This Question is Closed

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Not sure :p
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.2
dw:1350910904638:dw
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
That's also the point I reached. and then \(\rm \large x^2(x^{12 }1)  7 = 0\)
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
And then I factored it by difference of squares continuously.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Then I reached nothing.
 one year ago

ksaimouli Group TitleBest ResponseYou've already chosen the best response.2
dw:1350910926540:dw
 one year ago

benzilla Group TitleBest ResponseYou've already chosen the best response.0
X^14 x^2=7..........>X^2(X^121)...........>x^2(x1)(x^11+x^10.......) The second expression is prime so your two real roots are zero and one
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
\[x^{14}=7+x^2\]\[x^{14}x^2=7\]\[x^2(x^{12}1)=7\]
 one year ago

benzilla Group TitleBest ResponseYou've already chosen the best response.0
oops i made a mistake I forgot the 7
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
now what,
 one year ago

benzilla Group TitleBest ResponseYou've already chosen the best response.0
The equation is then xX2(x1)(x^11+x^10,,,,,,)=7
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
there are fourteen solutions
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.3
Just find real solutions.
 one year ago

benzilla Group TitleBest ResponseYou've already chosen the best response.0
I set this to zero which made this come out in error. This needs an iterative process to solve.
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.3
Yep... @benzilla
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
if there are some answers they are in the range\[\rm 1<x<2\]
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
answer will be close to 1 so letting\[x^2 \approx1\]\[x^{14}=8\]\[x=\sqrt[14]{8}\]
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
actually\[x\approx\pm\sqrt[14]{8}\]
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.3
What kind of solution is that???
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
thats just a approximate solution...not exact
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
we have\[x\approx\pm\sqrt[14]{8}=\pm1.16\]comparing this with exact solutions http://www.wolframalpha.com/input/?i=x%5E14%3D7%2Bx%5E2 its a good approaximate :)
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.3
Yep... What about the exact solution.
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.3
I think it is irrational
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
well thats what engineers do...lets think like mathematicians
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.3
Let y=x^2 then, y^7=7+y y=(7+y)^(1/7)
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.3
dw:1350912581733:dw
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
trial and error
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.3
And that gives:dw:1350912666885:dw
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.3
Thus, dw:1350912725583:dw
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.3
Can I do that?
 one year ago

benzilla Group TitleBest ResponseYou've already chosen the best response.0
Beyond a cubic function, there is not an equation like the quadratic. There are some relationships for some functions but what you have to do is yes as mukushla says trial and error. You take the number and keep plugging until you get close enough to the actual number. While we want a pure number, it will be hard to solve any other way.
 one year ago

ganeshie8 Group TitleBest ResponseYou've already chosen the best response.1
thats very much an exact solution :)
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.3
And @UnkleRhaukus from similar process the solution of x^3=1+x is dw:1350913011409:dw
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
hmm, good thinking
 one year ago

benzilla Group TitleBest ResponseYou've already chosen the best response.0
If you want to try to find all factors here is a little more to warp the brain. It is all factored except for two cubic terms. X^2(x^3+i)(x^3i) (x1)[X1/2[(sqrt3)/2]i] [X1/2+[(sqrt3)/2]i](x+1) [X+1/2+[(sqrt3)/2]i] [X1/2[(sqrt3)/2]i]=7 X^2(X+i)(?)(Xi)(??) (x1)[X1/2[(sqrt3)/2]i] [X1/2+[(sqrt3)/2]i](x+1) [X+1/2+[(sqrt3)/2]i] [X1/2[(sqrt3)/2]i]=7 The ? and ?? can be factored however i am running out of time this morning. Good luck
 one year ago

benzilla Group TitleBest ResponseYou've already chosen the best response.0
Here it is completely factored. I couldn't resist. A lot of fun but what a mess. You will see there are a lot of roots in this equation. X^2(X+i )(X +i/2 +[sqrt3]/2)(X+i/2[sgrt3]/2)(Xi)(Xi(1+sqrt3)/2)(Xi(1sqrt3)/2) (x1)[X+1/2(sqrt3)i] [X+1/2+(sqrt3)i](x+1) [X1/2+(sqrt3)i] [X1/2(sqrt3)i]=7 By zeros we know there are roots around +sqrt7 and sqrt7 and +1 (someone else found +1.16) and 1 which can be used to start your iteration. I went through this exercise to show how many roots there are in this equation. Good job on your work.
 one year ago

benzilla Group TitleBest ResponseYou've already chosen the best response.0
Oh sorry +1/7 instead of +1. See this makes the eyes blur. The best iterative program is to subtract the error if your over and determine a ratio multiplier that wont overshoot the answer. You can do this with a simple loop.
 one year ago

benzilla Group TitleBest ResponseYou've already chosen the best response.0
I meant over and under. It will converge to the answer.
 one year ago
See more questions >>>
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.