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14 Feb 2023, 02:30
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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:
50% (02:45) correct
50%
(02:47)
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What is the equation where roots are \((α + \frac{1}{β})\) and \((β + \frac{1}{α})\), if α and β are roots of \(ax^2 + bx + c = 0\)?
A. \(acx^2 + b(a + c)x + (a + c)^2 = 0\)
B. \(acx^2 + b(a + c)x - (a + c)^2 = 0\)
C. \(acx^2 - b(a + c)x + (a + c)^2 = 0\)
D. \(acx^2 - b(a + c)x - (a + c)^2 = 0\)
E. \(acx^2 + b(a + c)x + (a + c) = 0\)
700-Level question!!!
A. \(acx^2 + b(a + c)x + (a + c)^2 = 0\)
B. \(acx^2 + b(a + c)x - (a + c)^2 = 0\)
C. \(acx^2 - b(a + c)x + (a + c)^2 = 0\)
D. \(acx^2 - b(a + c)x - (a + c)^2 = 0\)
E. \(acx^2 + b(a + c)x + (a + c) = 0\)
700-Level question!!!
14 Feb 2023, 03:25
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As α and β are the roots of the equation \(ax^2+bx+c=0\)
sum of roots, α + β = \(\frac{-b}{a}\) and
product of roots αβ = \(\frac{c}{a}\)
We are required to find the equation whose roots are \(α + \frac{1}{β }\)and \(β + \frac{1}{α}\)
and the desired equation is -
\(x^2\) - (sum of roots)\(x\) + product of roots = 0
so sum of roots = \(α + β + \frac{1}{α} + \frac{1}{β} \)= \((α + β) + \frac{(α + β)}{αβ}\) = \(\frac{-b}{a}\) + (\(\frac{-b}{a}\))/(\(\frac{c}{a}\)) = \(\frac{-b}{a} - \frac{b}{c}\) = \(\frac{-b(c+a)}{ac}\)
product of roots = \(αβ + 2 + \frac{1}{αβ}\) = \(\frac{c}{a }+ 2 + \frac{a}{c}\) = \((c^2 + a^2)\)/ac + 2
substituting above, we get the desired equation as
\(x^2 - [-b\frac{(c+a)}{ac}]x + \frac{(c^2 + a^2)}{ac} + 2 = 0\)
\(acx^2 + b(c+a)x + (c^2 + a^2 + 2ac) = 0\\
acx^2 + b(c+a)x + (c+a)^2 = 0\)
Answer A
sum of roots, α + β = \(\frac{-b}{a}\) and
product of roots αβ = \(\frac{c}{a}\)
We are required to find the equation whose roots are \(α + \frac{1}{β }\)and \(β + \frac{1}{α}\)
and the desired equation is -
\(x^2\) - (sum of roots)\(x\) + product of roots = 0
so sum of roots = \(α + β + \frac{1}{α} + \frac{1}{β} \)= \((α + β) + \frac{(α + β)}{αβ}\) = \(\frac{-b}{a}\) + (\(\frac{-b}{a}\))/(\(\frac{c}{a}\)) = \(\frac{-b}{a} - \frac{b}{c}\) = \(\frac{-b(c+a)}{ac}\)
product of roots = \(αβ + 2 + \frac{1}{αβ}\) = \(\frac{c}{a }+ 2 + \frac{a}{c}\) = \((c^2 + a^2)\)/ac + 2
substituting above, we get the desired equation as
\(x^2 - [-b\frac{(c+a)}{ac}]x + \frac{(c^2 + a^2)}{ac} + 2 = 0\)
\(acx^2 + b(c+a)x + (c^2 + a^2 + 2ac) = 0\\
acx^2 + b(c+a)x + (c+a)^2 = 0\)
Answer A
15 Feb 2023, 15:01
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Let the equation be :
X^2+2X+1, whose roots are therefore -1,-1
That means the roots of the correct answer both equal (-1)-1=-2
That means the 2nd and 3rd terms of the correct answer must equal 4, or X^2+4X+4
Plugging the coefficients into answer A yields the above requirement.
X^2+2X+1, whose roots are therefore -1,-1
That means the roots of the correct answer both equal (-1)-1=-2
That means the 2nd and 3rd terms of the correct answer must equal 4, or X^2+4X+4
Plugging the coefficients into answer A yields the above requirement.
