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27 Feb 2023, 02:30
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A
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Difficulty:
65%
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Question Stats:
50% (02:07) correct
50%
(02:39)
wrong
based on 10
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What is the equation where roots are \((α + β)^2\) and \((α – β)^2\), if α and β are roots of \(x^2 + 3x + 4 = 0\) ?
A. \(x^2 - 2x - 63 = 0\)
B. \(x^2 - 2x + 63 = 0\)
C. \(x^2 + 2x - 63 = 0\)
D. \(x^2 + 2x + 63 = 0\)
E. \(x^2 - 2x + 65 = 0\)
700-Level question!!!
A. \(x^2 - 2x - 63 = 0\)
B. \(x^2 - 2x + 63 = 0\)
C. \(x^2 + 2x - 63 = 0\)
D. \(x^2 + 2x + 63 = 0\)
E. \(x^2 - 2x + 65 = 0\)
700-Level question!!!
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27 Feb 2023, 03:25
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Given that α and β are roots of x^2+3x+4=0
sum of roots (α+β) = -3 and product of roots (αβ) = 4
therefore, the desired equation whose roots are (α+β)^2 and (α–β)^2 is -
x2 - (sum of roots)x + product of roots = 0
sum of roots = 9 + (9 - 16) {as (α-β)^2 = α^2+β^2 -2αβ = α^2+β^2 + 2αβ - 4αβ = (α+β)^2 - 4αβ }
= 9 + (-7) = 2
product of roots = 9*(-7) = -63
the equation is x^2 -2x -63 = 0
Answer A
sum of roots (α+β) = -3 and product of roots (αβ) = 4
therefore, the desired equation whose roots are (α+β)^2 and (α–β)^2 is -
x2 - (sum of roots)x + product of roots = 0
sum of roots = 9 + (9 - 16) {as (α-β)^2 = α^2+β^2 -2αβ = α^2+β^2 + 2αβ - 4αβ = (α+β)^2 - 4αβ }
= 9 + (-7) = 2
product of roots = 9*(-7) = -63
the equation is x^2 -2x -63 = 0
Answer A
