Help in problem of theory of equations: Let a ≠ 0, b, c be integers and sinx, cosx be the rational roots of the equation ax^2+bx+c=0....
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What are your thoughts? What have you tried? How far have you gotten? ("Read Before Posting")
Please be complete. Thank you!
The Highlander
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\(\displaystyle sin\,x + cos\,x = -\frac{b}{a}\)
and
\(\displaystyle sin\,x\times cos\,x=\frac{c}{a}\)
Hint: Maybe start by squaring the former?
(That will remove the negative sign too.
)Please come back and show us how you have proceeded from there...
(Or show us your attempt at progress and say where you got stuck if you can't reach the final answer.)
Hope that helps.
BigBeachBanana
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Did you quote the problem exactly?
The question makes no sense if the arguments of [imath]\sin(x)[/imath] and [imath]\cos(x)[/imath] are [imath]x[/imath]. As written,
[imath](x-\sin x)(x-\cos x) = 0 \implies x =0[/imath] is one of the solution. When [imath]x = 0[/imath], this is a linear line; not a quadratic.
It would make sense if they were a different variable say [imath]\theta[/imath].
What is the QUESTION?
The Highlander
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The "QUESTION" is which of the given options (A to D) is true.
However, I too am wondering whether the problem has been quoted exactly?
Working through (from Post #3), a2 + 2ca is a perfect square but I don't see any of the given options being one!
(Perhaps Option B should have been "a2 + 2ca" rather than "a + 2c"?)
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Then that should have EXPLICITLY stated.....
BigBeachBanana
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Expand [imath](x-\sin \theta)(x-\cos \theta)[/imath] leads to one true conclusion.
For the answer, both sinΘ and cosΘ has to be rational - at some Θ. I see only Θ = n/2 * π (n = 0, 1, 2, 3, ......) meeting that constraint.
The Highlander
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You're right, of course, and the original statement should have been written as:-
"Let a (≠ 0), b, & c be integers and sin θ & cos θ be the rational roots of the equation ax2 + bx + c = 0."
and in my post (#3) I should have written:-
\(\displaystyle sin\,θ+cos\,θ=-\frac{b}{a}\)
and
\(\displaystyle sin\,θ\times cos\,θ=\frac{c}{a}\)
but I still don't get any of the given options as true!
I suspect the question has just been badly transcribed throughout.
Dr.Peterson
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I found one other source that suggests the problem is really
It is easy to see one case, where the roots are 0 and 1 (sine and cosine of 0); that results in the equation x^2 - x + 0 = 0, where a is a perfect square, b is the square of an imaginary number (does that count?), and a + 2c and a - 2c are perfect squares; but these may not always be true.
On the other hand, the equation could also be 2x^2 - 2x + 0 = 0, in which case none of the conclusions are true! So either another condition is required, or something else is copied incorrectly.
Now, I tried determining the condition under which, if p and q are the rational roots, it is possible that p^2 + q^2 = 1. I got something else being a perfect square. So possibly the problem just needs a little tweak.
(Until I realized the conditions of the problem are possible, I wondered if it might be a trick question such that all four conclusions are vacuously true ...)
Let a ≠ 0, b, c be integers and sinθ, cosθ be the rational roots of the equation ax^2+bx+c=0. Then [which of the following is true]:
(A) a is a perfect square
(B) a + 2c is a perfect square
(C) a - 2c is a perfect square
(D) b is a perfect square
It is easy to see one case, where the roots are 0 and 1 (sine and cosine of 0); that results in the equation x^2 - x + 0 = 0, where a is a perfect square, b is the square of an imaginary number (does that count?), and a + 2c and a - 2c are perfect squares; but these may not always be true.
On the other hand, the equation could also be 2x^2 - 2x + 0 = 0, in which case none of the conclusions are true! So either another condition is required, or something else is copied incorrectly.
Now, I tried determining the condition under which, if p and q are the rational roots, it is possible that p^2 + q^2 = 1. I got something else being a perfect square. So possibly the problem just needs a little tweak.
(Until I realized the conditions of the problem are possible, I wondered if it might be a trick question such that all four conclusions are vacuously true ...)
Hello guys I'm really sorry for not replying but I had my account linked to an email which I don't use and forgot I had asked this question. Anyways I saw that I had written cosx and sinx and yes it should be cos(theta) and sin(theta). And I ended up solving the question, as if a is an perfect square so will be a+2c and a-2c as well. Further I proved a must be a perfect square by assuming roots as p/q and √(1-(p/q)²)(since rational roots) from where we get that, P(x)=a(x-p/q)(x-√(1-(p/q)²)) on simplification and from there further deductions we can arrive on the answer as a, b, c
Dr.Peterson
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How do you deal with my observation (in #11) that 2x^2 - 2x + 0 = 0 has rational roots 0 = sin(0) and 1 = cos(0), but none of the options are true?
And on what grounds do you say "if a is a perfect square so will be a+2c and a-2c as well"?
Now, I can find an example of such a quadratic for which A, B, and C are true, but not D; but if you multiply it by a non-square, that will change.
Something is definitely not right here. I have to ask you again to show the exact wording of the original problem.
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The problem as stated expect for sinx it is sin(theta) is the problem word to word. I think the problem setter hadn't accounted for quadratics of the form you are talking about because even in my proof a must only be a multiple of a perfect square not a perfect square necessarily. Making the problem inherently wrong
blamocur
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The problem should include [imath]GCD(a,b,c) = 1[/imath].