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
 2 years ago
 2 years ago
sauravshakya Group Title
Another Fun Question Find all the real solutions: x^14=7+x^2
 2 years ago
 2 years ago

This Question is Closed

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

mukushla Group TitleBest ResponseYou've already chosen the best response.0
thats just a approximate solution...not exact
 2 years 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 :)
 2 years ago

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

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

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

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

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

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

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

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

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

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

benzilla Group TitleBest ResponseYou've already chosen the best response.0
I meant over and under. It will converge to the answer.
 2 years 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.