• Support PF! Buy your school textbooks, materials and every day products Here!

Perfect square

281
0
1. Homework Statement
[tex]\left(\frac{\dot{x}}{a}\right)^2 = K\left[b\frac{a}{x} + c\left(\frac{a}{x}\right)^2 + (1-b-c)\right][/tex]

Show that, for [itex]b<1[/itex], there is a value of [itex]c[/itex] that makes the right hand side a perfect square of a function of x.


2. The attempt at a solution
I guess that a perfect square is a square of a function that can be written as

[tex](x \pm d)^2 = x^2 \pm 2xd + d^2[/tex]

but I have not been successful in this problem.
 

Answers and Replies

AlephZero
Science Advisor
Homework Helper
6,953
291
That's the right general idea, but for your equation you want something like ((a/x) + d)^2.

If you have an expression like ax^2 + bx + c and it is a perfect square, what is special about the two roots of the quadratic equation ax^2 + bx + c = 0? What condition can you use to decide what kind of roots a quadratic equation has?
 
281
0
If [itex]ax^2+bx+c[/itex] is a perfect square, then [itex]b=2 \sqrt{ac}[/itex] and the roots of [itex]ax^2+bx+c=0[/itex] is a double root [itex]x_{1,2}=-\sqrt{c/a}[/itex]. I'm not sure what you mean with the conditions..
 
HallsofIvy
Science Advisor
Homework Helper
41,732
893
First let u= a/x. Now, what value of c makes cu2+ bu+ 1-b-c a perfect square? Also recall that Ax2+Bx+ C is a perfect square if and only if B2-4AC= 0. However, it seems to me that b would have to be less than or equal to 1/2 for this to be true, not just less than 1.
 
123
0
and if [itex]b^2=4ac[/itex] then [itex]ax^2+bx+c[/itex] is a perfect square. So you know the condition already.

Look again at your quadratic (shame it uses the same letters as those in the general form of the quadratic you have given)

In your quadratic what is equivalent to 'x' 'a', 'b', 'c' in the general form?
 
281
0
Now, what value of c makes cu2+ bu+ 1-b-c a perfect square?
Then

[tex]c = \frac{b-1 \pm \sqrt{1-2b}}{2}[/tex]

but that's two values. I think I also want to have [itex]-1 \leq c \leq 1[/itex] for my problem.
 
281
0
Ok, this is correct. But now I need to find an exact solution for [itex]x(t)[/itex] in this case. Help..? =)
 
HallsofIvy
Science Advisor
Homework Helper
41,732
893
Now write that right hand side as a perfect square, take the square root of both sides and solve the resulting separable differential equation.
 
281
0
Then I get, using [itex]u=a/x[/itex],

[tex]u=\sqrt{\frac{1-b-c}{c}} \left[ \frac{(exp[K(1-b-c)^{1/4}t]+1)^2}{(exp[K(1-b-c)^{1/4}t]-1)^2} - 1 \right][/tex]

and when I then try to solve for x and show that this is a solution I get these infinite calculations so this doesn't seem right.
 

Related Threads for: Perfect square

  • Last Post
Replies
11
Views
6K
  • Last Post
Replies
2
Views
6K
Replies
13
Views
9K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
1
Views
4K
Replies
6
Views
4K
Top