Solution of polynomial equation

Subhotosh Khan said:
In the post above, you did not tell us what is it that you need to find. Please tell us about those ideas - even if you know those are bad. Telling us will provide a starting point for our discussion.

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I need to find every a for which this equation Will have 4 distinct real roots, but this is not a quadratic equation, if it was then i would know, i have tried factorizing but i didnt get anywhere, I have no other hints.
 
BadMathematician said:
I need to find every a for which this equation Will have 4 distinct real roots, but this is not a quadratic equation, if it was then i would know, i have tried factorizing but i didnt get anywhere, I have no other hints.
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Please share your "work" - so that we can teach you fruitfully !!
 
BadMathematician said:
I need to find every a for which this equation will have 4 distinct real roots, but this is not a quadratic equation, if it was then i would know, i have tried factorizing but i didn't get anywhere, I have no other hints.
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I don't see a good way to solve this algebraically, since you ended up with a fourth-degree polynomial (unless you know the discriminant for quartics, which I haven't bothered to look up because I recall that it's pretty complicated).

So I'd be inclined to sketch the graph of the original function, [imath]x^2+\left(\frac{x}{x-1}\right)^2[/imath], using calculus techniques, which will easily determine the answer.
 
Dr.Peterson said:
I don't see a good way to solve this algebraically, since you ended up with a fourth-degree polynomial (unless you know the discriminant for quartics, which I haven't bothered to look up because I recall that it's pretty complicated).

So I'd be inclined to sketch the graph of the original function, [imath]x^2+\left(\frac{x}{x-1}\right)^2[/imath], using calculus techniques, which will easily determine the answer.
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I havent yet learned how to draw polynomial graphs I am still in highschool but when I put it in online graph maker, it looks Like two power graphs with x=1 asymptote, but I still dont know what that tells me about a.

Is it possible to rewrite it as a quadratic equation where coeffitients still contain variable x? I know when a quadratic has real distinct solutions but how am I supposed to solve this with my knowedge?
 
BadMathematician said:
I havent yet learned how to draw polynomial graphs I am still in highschool but when I put it in online graph maker, it looks Like two power graphs with x=1 asymptote, but I still dont know what that tells me about a.

Is it possible to rewrite it as a quadratic equation where coeffitients still contain variable x? I know when a quadratic has real distinct solutions but how am I supposed to solve this with my knowedge?
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I have no idea what you are expected to do! This is not a polynomial, but a rational function, with y = x^2 as a "curved asymptote", as well as the vertical asymptote.

But if you graph it, you can think about what horizontal lines y = a will intersect the graph in 4 places. Here is my graph, which I assume agrees with yours:

1638453375757.png​

Have you asked your teacher what you are expected to do?

And can you show us the actual problem as given to you, so we can see if it gives any clues about that? You started this thread with a totally inadequate statement of the problem, and I suspect there are details you still haven't told us.
 
The problem was exactly "determine every Real number a so that the given function has exactly 4 different real solutions" , I see what you mean now with that graph, the solutions would be a>8 but I dont know how to draw that graph.
 
BadMathematician said:
The problem was exactly "determine every Real number a so that the given function has exactly 4 different real solutions" , I see what you mean now with that graph, the solutions would be a>8 but I dont know how to draw that graph.
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The odd thing, is, a function doesn't have solutions! It's the equation they are talking about. So the problem is written incorrectly (not that that changes anything).

Again, you need to ask your teacher about the problem. That is allowed! There is nothing more I can say.
 
This was posted in Beginning Algebra. I am sure that Dr. Peterson is correct that use of the quartic formula could be used to get an answer by purely algebraic means. But the quartic formula is not beginning algebra. And Dr. Peterson‘s graphic method is very nice. But the OP wanted an analytic answer. A very simple one is available through calculus.

Consider the family of real functions specified by

[math]x \ne 1 \text { and } f(x) = y = x^2 + \left ( \dfrac{x}{x - 1} \right )^2 - a.[/math]
How can we describe the real roots of such a function, realizing that the roots will be the solutions to equations of the form

[math]x^2 + \left ( \dfrac{x}{x - 1} \right )^2 = a.[/math]
It should be obvious that if a is negative, the function will have no real roots.

What is the derivative of f(x)?

[math]\dfrac{dy}{dx} = 2x + 2x(x - 1)^{-2} + x^2(-2)(x - 1)^{-3} = 2x * \left ( 1 + \dfrac{x - 1}{(x - 1)^3} - \dfrac{x}{(x - 1)^3}\right ) =[/math]
[math]\dfrac{2x}{(x - 1)^3} * \left ( x^3 - 3x^2 + 3x - 2 \right ) = \dfrac{2x(x - 2)(x^2 - x + 1)}{(x - 1)^3.}[/math]
The term [math]x^2 - x + 1[/math] has no real roots so the derivative is equal to 0 only at x = 0 and x = 2.

If x < 0, the derivative is negative. If 0 < x < 1, the derivative is positive. Thus the function has a minimum at x = 0.

If 1 < x < 2, the derivative is negative. If x > 0, the derivative is positive. Thus the function has another minimum at x = 2.

At x = 0, the function equals - a. Thus, if a = 0, the function has a duplicated real root at x = 0. If a = 0, the function equals 8 at x = 2.

If a = 8, we have three real roots. If a > 8, there are four real roots.

Going back to the equation, we can summarize

If a is negative, there are no real solutions.

If a is zero, there is one real solution, namely zero.

0 < a < 8, there are two real solutions.

If a = 8, there are three real solutions, one at x = 2.

If a > 8, there are four real solutions.

If there is more than one real solution, one is negative and another is between 0 and 1.

If there are four solutions, one is between 1 and 2, and another exceeds 2.

Relatively easy calculus problem. Horrific algebra problem.
 
This, of course, is what I referred to in #7:
Dr.Peterson said:
I don't see a good way to solve this algebraically, since you ended up with a fourth-degree polynomial (unless you know the discriminant for quartics, which I haven't bothered to look up because I recall that it's pretty complicated).

So I'd be inclined to sketch the graph of the original function, [imath]x^2+\left(\frac{x}{x-1}\right)^2[/imath], using calculus techniques, which will easily determine the answer.
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I hope we can find out eventually what the teacher was thinking.
 
BadMathematician said:
I need to find every a for which this equation Will have 4 distinct real roots, but this is not a quadratic equation, if it was then i would know, i have tried factorizing but i didnt get anywhere, I have no other hints.
Click to expand...
You really thought that someone here would have known that is what you needed to find??
 
BadMathematician said:
The problem was exactly "determine every Real number a so that the given function has exactly 4 different real solutions" , I see what you mean now with that graph, the solutions would be a>8 but I dont know how to draw that graph.
Click to expand...
Every quartic polynomial can be factored into two quadratic functions.

ref: https://www.maa.org/sites/default/files/Brookfield2007-103574.pdf

From there you can deal with solutions of quadratic equations. But this involves tedious algebraic manipulations.
 
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