Nature of roots of an equation

[zorro]

Homework Statement

The equation A²/(x-a) + B²/(x-b) + C²/(x-c) + ... + H²/(x-h) = k has

1) No real root
2) At most one real root
3) No complex root
4) At most two complex roots

The Attempt at a Solution

We can see that the function f(x) = L.H.S. - R.H.S. is a strictly decreasing function with x=a,b,c...h as its asymptotes. This is all I could think of.

 

[jambaugh]

Hmmmm...

My first thought is to get a common denominator but that seems intractable. However it will allow you to determine the degree and thus count total complex roots via the Fundamental Theorem of Algebra.

Consider how the graph (of the L.H.S.) must look. You know each Vertical Asymptote is of odd multiplicity. How does the curve to the right of say x=a match up with the curve to the left of x=b? What is the range in this interval?

Look at how the graph of y=L.H.S. looks in terms of crossing the horizontal line y = k.

Finally I think the multiple choices are ill formed since 3 implies 4 but presumably you are to give the most specific choice. Also "real roots" are technically special cases of "complex roots" and thus technically 3) implies 1), and 4) implies 2). The author may have intended "complex root" to mean "complex non-real root". Otherwise, as I parse the literal meaning of the answers, I'd have to say "none of the above".

 

[Ray Vickson]

Homework Statement

The equation A²/(x-a) + B²/(x-b) + C²/(x-c) + ... + H²/(x-h) = k has

1) No real root
2) At most one real root
3) No complex root
4) At most two complex roots

The Attempt at a Solution

We can see that the function f(x) = L.H.S. - R.H.S. is a strictly decreasing function with x=a,b,c...h as its asymptotes. This is all I could think of.

Try plotting a simple example like f(x) = 1/(x-1) + 1/(x-2) + 1/(x-3) to see what happens.

RGV

 

[hunt_mat]

You know that this will lead to a polynomial of degree 8 right? Why? There are 8 letters from A to H inclusive. Look at the constant term, what does this tell you?

 

[SammyS]

Homework Statement

The equation A²/(x-a) + B²/(x-b) + C²/(x-c) + ... + H²/(x-h) = k has

1) No real root
2) At most one real root
3) No complex root
4) At most two complex roots

The Attempt at a Solution

We can see that the function f(x) = L.H.S. - R.H.S. is a strictly decreasing function with x=a,b,c...h as its asymptotes. This is all I could think of.

Rewrite the function you suggested, f(x) = A²/(x-a) + B²/(x-b) + C²/(x-c) + ... + H²/(x-h) - k, as a rational function. The horizontal asymptote is k. Between each neighboring pair of vertical asymptotes, this function ranges from -∞ to +∞ .

 

[zorro]

Look at the constant term, what does this tell you?

I can't figure that out.

Rewrite the function you suggested, f(x) = A²/(x-a) + B²/(x-b) + C²/(x-c) + ... + H²/(x-h) - k, as a rational function. The horizontal asymptote is k. Between each neighboring pair of vertical asymptotes, this function ranges from -∞ to +∞ .

How is k a horizontal asymptote? I agree that the function ranges from -∞ to +∞ but what does it say about the nature of roots?

 

[hunt_mat]

Put everything on the LHS over a common denominator, multiply by the common denominator and expand the RHS. The constant term will be abcdefghk

 

[SammyS]

How is k a horizontal asymptote? I agree that the function ranges from -∞ to +∞ but what does it say about the nature of roots?

I should have said -k.

lim _{x → ±∞} f(x) = -k

Also, if you use a common denominator (somewhat like jambaugh suggested) to rewrite f(x) as a rational function, you get:

$$f(x)=\frac{A^2(x-b)(x-c)\dots(x-h)+B^2(x-a)(x-c)\dots(x-h)+\dots+H^2(x-a)\dots(x-g)\ -\ k(x-a)\dots(x-g)(x-h)}{(x-a)(x-b)(x-c)(x-d)(x-e)(x-f)(x-g)(x-h)}$$

$$=\frac{-kx^8+(A^2+B^2+\dots+H^2+k(a+b+\dots+h)x^7+\dots}{(x-a)(x-b)(x-c)(x-d)(x-e)(x-f)(x-g)(x-h)}$$

$$=\frac{-kx^8+(A^2+B^2+\dots+H^2+k(a+b+\dots+h)x^7+\dots}{x^8-(a+b+\dots+h)x^7+\dots}$$​

Also, graph some simpler but similar function like g(x)=1/(x+2)+3/(x-2)+2/(x_4) - 3 .

 

[Stephen Tashi]

1) No real root
2) At most one real root
3) No complex root
4) At most two complex roots

This is an interesting question since it doesn't offer "5) None of the above". How do your materials define a "complex root". After all, real numbers are a subset of the complex numbers. And are we assuming all the constants in the equation are real numbers?

 

[Dick]

The problem only makes sense if you assume all of the constants are real. If you do then it makes good sense. The numerators are squared to ensure they are postive. Use Abdul Quadeer's observation that the function is always decreasing and sketch the asympotes. Use the graph to count the real roots for various cases of k. There really is a good answer. At least if you ignore some stupid phrasing problems in the A, B, C, D selections and just try to figure out what they want you to conclude.

 

[Stephen Tashi]

The problem only makes sense if you assume all of the constants are real.

.

Real and non-zero. And do we worry about possibilities like a = b?

This is the type of problem that tests one's mathematical skill and one's skill in reading the mind of the person who wrote it.


Abdul,
To summarize the advice so far, visualize adding the fractions together with a common denominator that is the product of the denominators. You get a numerator that is a polynomial of a certain degree. The function's zeroes are the zeroes of that polynomial. You can count the real zeroes of the function by imagining the graph of the original function. See if that accounts for all the zeroes of the function. Any not accounted for, must be complex.

 

[Dick]

.

Real and non-zero. And do we worry about possibilities like a = b?

Once you've read the mind of the question's author, you realize that those don't make any essential difference. If a=b you can combine those two into the same term. If any of the constants in the numerator are zero, you can drop them. The properly phrased answer is still true. The abject sloppiness of the question really hits you when you start trying to figure out the "At most" options. As SammyS has already pointed out.

 

[Stephen Tashi]

Once you've read the mind of the question's author, you realize that those don't make any essential difference.

That statement made me laugh, but I'll have to agree with it.

 

[zorro]

Suppose z is a root of the equation, so is its conjugate. I will denote it by z'.
Then we have

A²/(z-a) + B²/(z-b) + C²/(z-c) + ... + H²/(z-h) = k

and

A²/(z'-a) + B²/(z'-b) + C²/(z'-c) + ... + H²/(z'-h) = k

Let z - z' = iR, where R is a real number.

Subtracting the two equations and simplifying, we get

iR(A²/|z-a|² + B²/|z-b|² + C²/|z-c|² + ... + H²/|z-h|²) = 0

The bracketed term is positive (after assuming A,B,C... a,b,c are real). So we must have R=0. In other words, all roots are real/no complex root.

A.Q.

 

[Dick]

Suppose z is a root of the equation, so is its conjugate. I will denote it by z'.
Then we have

A²/(z-a) + B²/(z-b) + C²/(z-c) + ... + H²/(z-h) = k

and

A²/(z'-a) + B²/(z'-b) + C²/(z'-c) + ... + H²/(z'-h) = k

Let z - z' = iR, where R is a real number.

Subtracting the two equations and simplifying, we get

iR(A²/|z-a|² + B²/|z-b|² + C²/|z-c|² + ... + H²/|z-h|²) = 0

The bracketed term is positive (after assuming A,B,C... a,b,c are real). So we must have R=0. In other words, all roots are real/no complex root.

A.Q.

Sure, or you can see it from your asymptotes. If you clear the fractions to get a polynomial to solve and n is the number of terms, the polynomial has degree n if k is not zero and degree n-1 if k is zero. Now if you count the intersections of the graph with the line y=k you get n intersections if k is not zero and n-1 if k is zero. So the number of real roots is equal to the degree of the polynomial and there are no complex roots.

 

[zorro]

Sure, or you can see it from your asymptotes. If you clear the fractions to get a polynomial to solve and n is the number of terms, the polynomial has degree n if k is not zero and degree n-1 if k is zero. Now if you count the intersections of the graph with the line y=k you get n intersections if k is not zero and n-1 if k is zero. So the number of real roots is equal to the degree of the polynomial and there are no complex roots.

I think that will be the quickest method to get the answer without writing anything.