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13 Jul 2023, 01:30
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
48% (02:58) correct
52%
(02:38)
wrong
based on 77
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What is the equation whose roots are α and β, if \((\frac{-α^2}{β})\) and \((\frac{-β^2}{α})\) are roots of the equation \(3x^2 – 18x + 2 = 0\)?
A. \(3x^2 + 6x + 2 = 0\)
B. \(3x^2 - 6x + 2 = 0\)
C. \(3x^2 + 6x - 2 = 0\)
D. \(3x^2 - 6x - 2 = 0\)
E. \(3x^2 - 6x - 3 = 0\)
A. \(3x^2 + 6x + 2 = 0\)
B. \(3x^2 - 6x + 2 = 0\)
C. \(3x^2 + 6x - 2 = 0\)
D. \(3x^2 - 6x - 2 = 0\)
E. \(3x^2 - 6x - 3 = 0\)
13 Jul 2023, 04:47
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let our equation based on \(\alpha\) and \(\beta\) be :
\(x^2 - (\alpha + \beta)x + \alpha\beta = 0\) -------------------- 1)
We also know that \((-\alpha^{2}/\beta)\) and \((-\beta^{2}/\alpha)\) --------- a)
are roots of the equation \(3x^2–18x+2=0\)
from a) we can conclude that :
Firstly, multiplying them we get, \(\alpha\beta\) = \(\frac{c}{a}\) = \(\frac{2}{3}\).
Secondly adding them we get \(\frac{-\alpha^{2}}{\beta}+\frac{-\beta^{2}}{\alpha} = 6\) => \(\frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha} = -6\).
OR
\(\frac{\alpha^{3}+\beta^{3}}{\alpha\beta} = -6\) => substituting the value of \(\alpha\beta\) we get :
\(\alpha^{3}+\beta^{3}= -4\)
here use identity to convert the following into :
\((\alpha+\beta)^{3} - 3*\alpha\beta(\alpha+\beta) = -4\)
substitute value of \(\alpha+\beta=t\) and \(\alpha\beta]\) we get :
\(t^{3} - 2t + 4 = 0.\)
take a glance at the options available to simplify the calculation.
from that we can guess that t is either 2 or -2.
t = -2 satisfies the equation. hence
\( \alpha + \beta = -2\)
Now that we know \(\alpha+\beta\) and \(\alpha\beta\) we can put values in 1) to get
\(3x^{2} + 6x + 2 = 0\)
So IMO A
\(x^2 - (\alpha + \beta)x + \alpha\beta = 0\) -------------------- 1)
We also know that \((-\alpha^{2}/\beta)\) and \((-\beta^{2}/\alpha)\) --------- a)
are roots of the equation \(3x^2–18x+2=0\)
from a) we can conclude that :
Firstly, multiplying them we get, \(\alpha\beta\) = \(\frac{c}{a}\) = \(\frac{2}{3}\).
Secondly adding them we get \(\frac{-\alpha^{2}}{\beta}+\frac{-\beta^{2}}{\alpha} = 6\) => \(\frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha} = -6\).
OR
\(\frac{\alpha^{3}+\beta^{3}}{\alpha\beta} = -6\) => substituting the value of \(\alpha\beta\) we get :
\(\alpha^{3}+\beta^{3}= -4\)
here use identity to convert the following into :
\((\alpha+\beta)^{3} - 3*\alpha\beta(\alpha+\beta) = -4\)
substitute value of \(\alpha+\beta=t\) and \(\alpha\beta]\) we get :
\(t^{3} - 2t + 4 = 0.\)
take a glance at the options available to simplify the calculation.
from that we can guess that t is either 2 or -2.
t = -2 satisfies the equation. hence
\( \alpha + \beta = -2\)
Now that we know \(\alpha+\beta\) and \(\alpha\beta\) we can put values in 1) to get
\(3x^{2} + 6x + 2 = 0\)
So IMO A
23 Mar 2024, 09:17
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I believe there is an error somewhere. The math ain't mathing. I could be misguided in my math here.
For simplicity, I will refer alpha as 'a' and beta as 'b'.
From equation 2:
a) Product:
\(\frac{2}{3} = (\frac{-a^2}{b})(\frac{-b^2}{a})\)
\(\frac{2}{3} = -a*-b\)
\(2 = 3ab\)
\(4 = 6ab\) ----a
b) Sum:
\(\frac{18}{3} = (\frac{-a^2}{b}) + (\frac{-b^2}{a})\)
\(6 = \frac{(-a^3 - b^3)}{ab}\)
\(6ab = -a^3 - b^3\) (substitute the value of 6ab)
\(4 = -a^3 -b^3\) ----b
The Rationale:
From eq a, for 6ab to result in positive number, either both a & b must be negative or both positive. Now eq b helps us narrow down further: either both a^3 and b^3 must be negative to give a positive sum, OR the larger one of them should be negative.
So using eq a and b together, we get to know that both a (alpha) and b (beta) must be negative.
However, the answer to this question requires both to be positive (as the sum and product are both positive in the equation). Which goes against the equation we have derived.
I could be wrong here. Bunuel please help us.
For simplicity, I will refer alpha as 'a' and beta as 'b'.
From equation 2:
a) Product:
\(\frac{2}{3} = (\frac{-a^2}{b})(\frac{-b^2}{a})\)
\(\frac{2}{3} = -a*-b\)
\(2 = 3ab\)
\(4 = 6ab\) ----a
b) Sum:
\(\frac{18}{3} = (\frac{-a^2}{b}) + (\frac{-b^2}{a})\)
\(6 = \frac{(-a^3 - b^3)}{ab}\)
\(6ab = -a^3 - b^3\) (substitute the value of 6ab)
\(4 = -a^3 -b^3\) ----b
The Rationale:
From eq a, for 6ab to result in positive number, either both a & b must be negative or both positive. Now eq b helps us narrow down further: either both a^3 and b^3 must be negative to give a positive sum, OR the larger one of them should be negative.
So using eq a and b together, we get to know that both a (alpha) and b (beta) must be negative.
However, the answer to this question requires both to be positive (as the sum and product are both positive in the equation). Which goes against the equation we have derived.
I could be wrong here. Bunuel please help us.
