Let a ≠ 0, b, c be integers and sinx, cosx be the rational roots of the equation ax^2+bx+c=0. Then:
(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
\(\displaystyle sin\,x + cos\,x = -\frac{b}{a}\)Let a ≠ 0, b, c be integers and sinx, cosx be the rational roots of the equation ax^2+bx+c=0. Then:
(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
Did you quote the problem exactly?Let a ≠ 0, b, c be integers and sinx, cosx be the rational roots of the equation ax^2+bx+c=0. Then:
(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
What is the QUESTION?Let a ≠ 0, b, c be integers and sinx, cosx be the rational roots of the equation ax^2+bx+c=0. Then:
(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
The "QUESTION" is which of the given options (A to D) is true.What is the QUESTION?
Then that should have EXPLICITLY stated.....The "QUESTION" is which of the given options (A to D) is true.
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.and sinΘ, cosΘ be the rational roots of
You're right, of course, and the original statement should have been written as:-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].
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?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
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 wrongHow 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.
The problem should include [imath]GCD(a,b,c) = 1[/imath].in my proof a must only be a multiple of a perfect square not a perfect square necessarily. Making the problem inherently wrong