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23 Mar 2021, 06:30
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E
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Difficulty:
75%
(hard)
Question Stats:
61% (02:43) correct
39%
(02:29)
wrong
based on 178
sessions
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In the equation \(y^2 – py + q = 0\), y is a variable and p and q are constant. If the roots of the given equation are a and b, which of the following is the equation whose roots are \((ab + a + b)\) and \((ab – a – b)\) ?
A \(y^2 – 4y + p^2 = 0\)
B \(y^2 + yq^2 – p^2 = 0\)
C \(y^2 – 2y + q^2 + p^2 = 0\)
D \(y^2 + 2qy – q^2 – p^2 = 0\)
E \(y^2 – 2qy + q^2 – p^2 = 0\)
A \(y^2 – 4y + p^2 = 0\)
B \(y^2 + yq^2 – p^2 = 0\)
C \(y^2 – 2y + q^2 + p^2 = 0\)
D \(y^2 + 2qy – q^2 – p^2 = 0\)
E \(y^2 – 2qy + q^2 – p^2 = 0\)
ananya3
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Joined: 05 Oct 2018
Last visit: 17 Dec 2022
Posts: 106
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Location: India
Schools: Fuqua '24 Haas '24 ISB '23 Johnson '24 Kelley '24 Kenan-Flagler '24 McDonough '24 NUS Ross '24 Tepper '24 Tuck '24 Rotman '24 (A)
GMAT 1: 700 Q49 V36
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Schools: Fuqua '24 Haas '24 ISB '23 Johnson '24 Kelley '24 Kenan-Flagler '24 McDonough '24 NUS Ross '24 Tepper '24 Tuck '24 Rotman '24 (A)
GMAT 1: 700 Q49 V36
Posts: 106
23 Mar 2021, 11:06
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\( y^2–py+q=0\)
roots of the equation: a & b
=> a+ b = -b/a = -(-p)
=> a * b = c/a = q
new roots: (ab+a+b) and (ab–a–b)
=> -b / a = (ab + (a+b)) + (ab - (a+b))
=> -b / a = 2ab = 2q
=> c / a = (ab + (a+b)) * (ab - (a+b))
=> c / a = \((ab^2 - (a+b)^2)\) = \((q^2 - p^2)\)
=> equation
\(y^2 - 2yq + q^2 - p^2\)
IMO E
roots of the equation: a & b
=> a+ b = -b/a = -(-p)
=> a * b = c/a = q
new roots: (ab+a+b) and (ab–a–b)
=> -b / a = (ab + (a+b)) + (ab - (a+b))
=> -b / a = 2ab = 2q
=> c / a = (ab + (a+b)) * (ab - (a+b))
=> c / a = \((ab^2 - (a+b)^2)\) = \((q^2 - p^2)\)
=> equation
\(y^2 - 2yq + q^2 - p^2\)
IMO E
General Discussion
23 Mar 2021, 07:28
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a+b = p and ab = q
substitute the new roots
y^2-(ab+a+b+ab-a-b)y+{(ab+(a+b)) (ab-(a+b))}
cancel and substitute from above
y^2-2qy+(q^2-p^2)
Ans is E
substitute the new roots
y^2-(ab+a+b+ab-a-b)y+{(ab+(a+b)) (ab-(a+b))}
cancel and substitute from above
y^2-2qy+(q^2-p^2)
Ans is E
