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giokrutoi
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Homework Statement
x^2 - 5x -4(x)^1/2 +13=0
how to solve this
Homework Equations
The Attempt at a Solution
I couldn't figure out anithing
giokrutoi said:Homework Statement
##x^2 - 5x -4\sqrt{x} +13=0##
how to solve this
Homework Equations
The Attempt at a Solution
I couldn't figure out anithing
Cause there are no real roots. Sorry I just noticed.giokrutoi said:y^4 - 5 y^2 - 4y +13 = 0
but I can't find root
Fundamental theorem of algebra.giokrutoi said:so how can I count without solving
Yes.giokrutoi said:so it is two I guess
Welcomegiokrutoi said:thanks
giokrutoi said:Homework Statement
x^2 - 5x -4(x)^1/2 +13=0
how to solve this
Homework Equations
The Attempt at a Solution
I couldn't figure out anithing
giokrutoi said:so it is two I guess
I think the total number of roots are two, https://www.wolframalpha.com/input/?i=x^2+-+5x+-4(x)^(1/2)++13=0 .Ray Vickson said:No: your equation y^4 - 5 y^2 - 4y +13 = 0 is of degree 4, so has 4 roots, not just 2. Then (as can be checked explicitly) that produces 4 different roots x = y^2.
You cannot use the fact that your x-equation has highest power x^2, because it also contains √x and so is not a polynomial equation at all. The root-counting business based on the highest power applies ONLY to polynomials.
Buffu said:I think the total number of roots are two, https://www.wolframalpha.com/input/?i=x^2+-+5x+-4(x)^(1/2)++13=0 .
Did I miss something ?
So how will tell exact number of roots of this polynomial without finding ?PeroK said:If you let ##y = \sqrt{x}## then you get a 4th power polynomial in ##y##, which has 4 complex roots. But, not all these roots may be expressed as ##\sqrt{x}## for some ##x##. For example, take the polynomial:
##x^2 -25x +60 \sqrt{x} -36##
With ##y = \sqrt{x}## this becomes:
##y^4 -25y^2 +60y -36##
This has four real roots: ##y = 1, 2, 3, -6##
Only ##-6## is not a square root, so the original equation has solutions: ##x = 1, 4, 9##
Buffu said:So how will tell exact number of roots of this polynomial without finding ?
It at most have 4 but is there a way to get exact number ?
ok I understood that the main thing is that I encountered this problem on quiz in internet and there was no valuable thing I could do so I stated that I couldn't figure out anything but I guess the only solution will be to graph equation .epenguin said:You said at first you had no idea what to do.
Does this mean the problem has no relation you can see with anything you have been doing in your course? If you tell us what do you have been doing recently that would be one clue. With intractable, ugly and unsatisfactory questions like this, asking the student if he has copied the question/problem out right has often led to simplifications.
giokrutoi said:ok I understood that the main thing is that I encountered this problem on quiz in internet and there was no valuable thing I could do so I stated that I couldn't figure out anything but I guess the only solution will be to graph equation .
giokrutoi said:ok I understood that the main thing is that I encountered this problem on quiz in internet and there was no valuable thing I could do so I stated that I couldn't figure out anything but I guess the only solution will be to graph equation .
There are several methods for solving a quartic equation without actually solving it algebraically. One method is to use the Rational Root Theorem and test potential rational roots of the equation. Another method is to use the Descartes' Rule of Signs to determine the number of positive and negative roots. Additionally, the graph of the equation can be used to approximate the roots.
Yes, it is possible to solve a quartic equation without using the quadratic formula. The quartic formula can be used, although it is much more complex than the quadratic formula. As mentioned before, other methods such as the Rational Root Theorem and Descartes' Rule of Signs can also be used.
The advantage of solving a quartic equation without actually solving it is that it can save time and effort. Solving a quartic equation algebraically can be a complex and time-consuming process, but using other methods can provide a quicker solution. Additionally, it can also give an approximate solution if the equation has no rational roots.
No, not all quartic equations can be solved without actually solving them. Some quartic equations may have irrational or complex roots, which cannot be found using the Rational Root Theorem or Descartes' Rule of Signs. In these cases, solving the equation algebraically may be necessary.
Yes, there are real-life applications for solving quartic equations without actually solving them. For example, in engineering and physics, quartic equations can represent motion and force equations, and finding approximate solutions can help in predicting the behavior of a system. Additionally, in finance and economics, quartic equations can be used to model market trends and make predictions.