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In the equation y^2 – py + q = 0, y is a variable and p and q are cons

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Bunuel
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In the equation y^2 – py + q = 0, y is a variable and p and q are cons [#permalink] New post   23 Mar 2021, 06:30
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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 \(y^2 – 2qy + q^2 – p^2 = 0\)
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Sunil2424
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Re: In the equation y^2 – py + q = 0, y is a variable and p and q are cons [#permalink] New post   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
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Re: In the equation y^2 – py + q = 0, y is a variable and p and q are cons [#permalink] New post   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
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Re: In the equation y^2 py + q = 0, y is a variable and p and q are cons [#permalink] New post   22 Aug 2022, 22:52
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Bunuel wrote:
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 \(y^2 – 2qy + q^2 – p^2 = 0\)


The easiest would be to put p = 2 and q = 1 to get the roots as 1 and 1 (it becomes a perfect square). So, a = 1, b = 1
(ab + a + b) = 3
(ab – a – b) = -1
Sum of roots = 2 and product of roots = -3.
So equation would be \(y^2 - 2y - 3 = 0\) When we put p = 2 and q = 1 in options, we should get this equation.
To get constant term as -3, we will need p^2 and q/q^2 with opposite signs. Only option (E) has that so check it.

Answer (E)
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Re: In the equation y^2 py + q = 0, y is a variable and p and q are cons [#permalink] New post   23 Aug 2022, 00:40
Given: In the equation \(y^2 – py + q = 0\), y is a variable and p and q are constant.

Asked: 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 \((ab – a – b)\) ?

Since a & b are roots of the equation \(y^2 – py + q = 0\), y is a variable and p and q are constant.
a + b = p
ab = q

For the equation whose roots are \((ab + a + b)\) and \((ab – a – b)\)
Sum of roots = (ab + a + b) + (ab – a – b) = 2ab = 2q
Product of roots = (ab + a + b) (ab – a – b) = a^2b^2 - (a+b)^2 = q^2 - p^2

Equation: y^2 - 2qy + (q^2 -p^2) = 0

IMO E
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Re: In the equation y^2 py + q = 0, y is a variable and p and q are cons [#permalink] New post   24 Aug 2022, 03:07
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
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Re: In the equation y^2 py + q = 0, y is a variable and p and q are cons [#permalink]
24 Aug 2022, 03:07
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