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20 Feb 2023, 02:30
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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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(hard)
Question Stats:
62% (02:53) correct
38%
(03:00)
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based on 100
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What is the equation where roots are \((2α + \frac{1}{β})\) and \((2β + \frac{1}{α})\), if α and β are roots of \(2x^2 - 3x -5 = 0\)?
A. \(5x^2 - 12x - 32 = 0\)
B. \(5x^2 - 12x + 32 = 0\)
C. \(5x^2 + 12x - 32 = 0\)
D. \(5x^2 + 12x + 32 = 0\)
E. \(5x^2 - 12x - 16 = 0\)
A. \(5x^2 - 12x - 32 = 0\)
B. \(5x^2 - 12x + 32 = 0\)
C. \(5x^2 + 12x - 32 = 0\)
D. \(5x^2 + 12x + 32 = 0\)
E. \(5x^2 - 12x - 16 = 0\)
20 Feb 2023, 10:39
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What is the equation where roots are (\(2α+\frac{1}{β}\)) and (\(2β+\frac{1}{α}\)), if α and β are roots of \(2x^2−3x−5=0\)?
Eqn can be rewritten as \(2x^2−3x−5=0\)
=> \(x^2−(\frac{3}{2})x−(\frac{5}{2})=0\)
Sum of roots: α + β = \(-(-\frac{3}{2})\) =\(\frac{3}{2}\)
Product of roots: αβ = -(5/2)
We are required to find the equation whose roots are \(2α + \frac{1}{β }\)and \(2β + \frac{1}{α}\)
and the desired equation is -
\(x^2\) - (sum of roots)\(x\) + product of roots = 0
sum of roots = \(2(α + β) + \frac{1}{β} + \frac{1}{α}\)
=> \(2(\frac{3}{2}) + \frac{(α + β)}{αβ}\)
=> 3 - 3/5 = 12/5
product of roots = \((2α + \frac{1}{β})(2β + \frac{1}{α})\\
=> 4αβ +2 +2 + \frac{1}{αβ}\\
=> -10 +4 -\frac{2}{5} = -32/5\)
Therefore, the desired equation is
\(x^2 - (\frac{12}{5})x -\frac{32}{5} = 0\)
=> \(5x^2 -12x -32 =0\)
Answer A
Eqn can be rewritten as \(2x^2−3x−5=0\)
=> \(x^2−(\frac{3}{2})x−(\frac{5}{2})=0\)
Sum of roots: α + β = \(-(-\frac{3}{2})\) =\(\frac{3}{2}\)
Product of roots: αβ = -(5/2)
We are required to find the equation whose roots are \(2α + \frac{1}{β }\)and \(2β + \frac{1}{α}\)
and the desired equation is -
\(x^2\) - (sum of roots)\(x\) + product of roots = 0
sum of roots = \(2(α + β) + \frac{1}{β} + \frac{1}{α}\)
=> \(2(\frac{3}{2}) + \frac{(α + β)}{αβ}\)
=> 3 - 3/5 = 12/5
product of roots = \((2α + \frac{1}{β})(2β + \frac{1}{α})\\
=> 4αβ +2 +2 + \frac{1}{αβ}\\
=> -10 +4 -\frac{2}{5} = -32/5\)
Therefore, the desired equation is
\(x^2 - (\frac{12}{5})x -\frac{32}{5} = 0\)
=> \(5x^2 -12x -32 =0\)
Answer A
02 Mar 2023, 00:58
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Bunuel
What is the equation where roots are \((2α + \frac{1}{β})\) and \((2β + \frac{1}{α})\), if α and β are roots of \(2x^2 - 3x -5 = 0\)?
A. \(5x^2 - 12x - 32 = 0\)
B. \(5x^2 - 12x + 32 = 0\)
C. \(5x^2 + 12x - 32 = 0\)
D. \(5x^2 + 12x + 32 = 0\)
E. \(5x^2 - 12x - 16 = 0\)
Solution: A. \(5x^2 - 12x - 32 = 0\)
B. \(5x^2 - 12x + 32 = 0\)
C. \(5x^2 + 12x - 32 = 0\)
D. \(5x^2 + 12x + 32 = 0\)
E. \(5x^2 - 12x - 16 = 0\)
- α and β are roots of \(2x^2 - 3x -5 = 0\)
- This means \(α+β=-(\frac{-3}{2})=\frac{3}{2}\) and \(αβ=-\frac{5}{2}\)
- Equation with given roots is \(x^2-Sx+P=0\) where S is sum of the roots and P is the product
- Therefore, equation with roots \((2α + \frac{1}{β})\) and \((2β + \frac{1}{α})\) is \(x^2-(2α + \frac{1}{β}+2β + \frac{1}{α})x+(2α + \frac{1}{β})(2β + \frac{1}{α})=0\)
\(⇒x^2-(2(α+β)+\frac{1}{α}+\frac{1}{β})x+(4αβ+2+2+\frac{1}{αβ})=0\)
\(⇒x^2-(2\times \frac{3}{2}+\frac{α+β}{αβ})x+(4\times \frac{-5}{2}+4+\frac{1}{\frac{-5}{2}})=0\)
\(⇒x^2-(3+\frac{\frac{3}{2}}{\frac{-5}{2}})x+(-10+4-\frac{2}{5})=0\)
\(⇒x^2-(3-\frac{3}{5})x+(-6-\frac{2}{5})=0\)
\(⇒x^2-\frac{12x}{5}-\frac{32}{5}=0\)
\(⇒5x^2-12x-32=0\)
Hence the right answer is Option A
