Help finding solutions to sqrt(a+sqrt(-b))+sqrt(a-sqrt(-b))=4 and

  • Thread starter Thread starter andrewkg
  • Start date Start date
AI Thread Summary
The discussion focuses on solving two mathematical problems: finding integers that satisfy the equation sqrt(a+sqrt(-b)) + sqrt(a-sqrt(-b)) = 4 and determining the roots of the quadratic equation x^2 - x - (2 + 4 + 6 + ... + 2014). Participants explore simplifications and substitutions but struggle with cancellations leading to trivial results. One participant suggests using the quadratic formula for the second problem and emphasizes the importance of detailing thought processes in problem-solving. The conversation encourages deeper exploration of the relationships between variables and the nature of solutions, particularly in complex scenarios.
andrewkg
Messages
86
Reaction score
0

Homework Statement


Q A
Find all integers which satisfy sqrt(a+sqrt(-b))+sqrt(a-sqrt(-b))=4
Q B
Find all the roots for the quadratic equation x^2-x-(2+4+6+...+2014)


Homework Equations



---

The Attempt at a Solution


Well with both I keep coming up with 0=0 and cannot seem to get past that
 
Physics news on Phys.org
How are you attempting the problems?
 
In the first I tried simplifying and got a=8, but sqrt(-b) always canceled out. For the second I tried substituting for x or x^2 and it always canceled all the other values out.
 
Well, for the second one, what are you getting for the summation of 2 to 2014?
The roots I get for your second problem are x = 1008 and x = -1007

I get 1,015,056 for the summation of positive integers from 2 to 2014.
Then the problem becomes nothing more than factoring this:
x² - x - 1,015,056
 
andrewkg said:
In the first I tried simplifying and got a=8, but sqrt(-b) always canceled out. For the second I tried substituting for x or x^2 and it always canceled all the other values out.

You seem to be reluctant to show how you are thinking about the problems ... you say "I tried simplifying..." but don't tell us the simplification, you say "I tried substituting..." but do not tell us the substitution.

For the second one - there is a formula for finding the roots of any quadratic... why not use it?
(hint: put c=that long sum on the end and find the roots wrt c).

For the first one - you appear t be saying that you get a=8 and b can be anything... did you check this by putting a=8 and b=0 into the equation? What happened?

... may should try exploring the details a bit to get a feel for how it behaves.
i.e. the relation could be thought of as x + y = 4.
if x and y are real numbers, and x=y, what does that mean for a and b?
what happens is x and y are complex numbers and x=y?
You should find these situations don't cancel out... are there any more?
Does a systematic way of finding the solutions occur to you?

Overall - these problems must be set in the context of some lessons: what were the lessons about?
 
Last edited:
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top