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07 Sep 2020, 02:23
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E
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:
66% (02:33) correct
34%
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Consider the equation y^2 – py + q. If the roots of the given equation are a and b, find the equation whose roots are (ab + a + b) and (ab – a – b).
(A) y^2 – 4y + p^2= 0
(B) y^2 + yq2 – 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 + yq2 – 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
07 Sep 2020, 05:56
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In any quadratic equation of the form \(y = ax^2 + bx + c\), having roots \(\alpha\) and \(\beta\)
Sum of the roots = \(\alpha \space + \space \beta \) = \(-\frac{b}{a}\) and
Product of the roots = \(\alpha \beta\) = \(\frac{c}{a}\)
For the quadratic equation \(y^2 -py + q\) having roots a and b, then
a + b = \(-\frac{-p}{1} = p\) and
ab = \(\frac{q}{1}= q\)
To find the quadratic equation whose roots are (ab + a + b) = ab + (a + b) = q + p and (ab - a - b) = ab - (a + b) = q - p
Sum of roots = q + p + q - p = 2q
Product of the roots = (p + q) (p - q) = \(q^2 - p^2\)
In terms of the sum and product of roots, the quadratic equation can be written as \(y^2 - (Sum \space of \space roots)y + (Product \space of \space roots)\)
The required Equation is \(y^2 - 2qy + q^2 - p^2 = 0\)
Option E
Arun Kumar
Sum of the roots = \(\alpha \space + \space \beta \) = \(-\frac{b}{a}\) and
Product of the roots = \(\alpha \beta\) = \(\frac{c}{a}\)
For the quadratic equation \(y^2 -py + q\) having roots a and b, then
a + b = \(-\frac{-p}{1} = p\) and
ab = \(\frac{q}{1}= q\)
To find the quadratic equation whose roots are (ab + a + b) = ab + (a + b) = q + p and (ab - a - b) = ab - (a + b) = q - p
Sum of roots = q + p + q - p = 2q
Product of the roots = (p + q) (p - q) = \(q^2 - p^2\)
In terms of the sum and product of roots, the quadratic equation can be written as \(y^2 - (Sum \space of \space roots)y + (Product \space of \space roots)\)
The required Equation is \(y^2 - 2qy + q^2 - p^2 = 0\)
Option E
Arun Kumar
General Discussion
yashikaaggarwal
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07 Sep 2020, 03:49
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Given: y^2 – py + q. The roots are a and b,
=> -b/a = sum of roots => P =A+B --------------------(1)
=> c/a = product of roots => Q = AB --------------------(2)
To find: find the equation whose roots are (ab + a + b) and (ab – a – b).
Solution :
Sum : AB+A+B+AB-A-B = 2AB = R.H.S
Product = (AB+A+B)*(AB-A-B)
=> (AB)^2 + A^2*B + A*B^2 + A^2*B - A^2 - AB + AB^2 -AB - B^2
=> 2*A^2*B + 2*A*B^2 + (AB)^2 - 2AB - A^2 - B^2 (Product R.H.S)
A)y^2 – 4y + p^2= 0
Sum = 4 = root1+root2 ( Not equal to R.H.S)
B)y^2 + yq2 – p^2 = 0
Sum = Q^2 = root1+root2
=> (AB)^2 ( Not equal to R.H.S)
C)y^2 – 2y + q^2 + p^2 = 0
Sum = root1+root2
=> 2 ( Not equal to R.H.S)
D)y^2 + 2qy – q^2 – p^2 = 0
Sum = 2Q = root1+root2
2AB (equal to SUM R.H.S)
Product = -Q^2-P^2
=> -AB^2-A^2-B^2-2AB (Not equal to Product R.H.S)
E)y^2 – 2qy + q^2 – p^2 = 0
Sum = 2Q = root1+root2
=> 2AB (equal to SUM R.H.S)
Product = +Q^2-P^2 = (Q+P)(Q-P)
=> (AB+A+B)(AB-A-B) (equal to Product R.H.S)
Answer is E
=> -b/a = sum of roots => P =A+B --------------------(1)
=> c/a = product of roots => Q = AB --------------------(2)
To find: find the equation whose roots are (ab + a + b) and (ab – a – b).
Solution :
Sum : AB+A+B+AB-A-B = 2AB = R.H.S
Product = (AB+A+B)*(AB-A-B)
=> (AB)^2 + A^2*B + A*B^2 + A^2*B - A^2 - AB + AB^2 -AB - B^2
=> 2*A^2*B + 2*A*B^2 + (AB)^2 - 2AB - A^2 - B^2 (Product R.H.S)
A)y^2 – 4y + p^2= 0
Sum = 4 = root1+root2 ( Not equal to R.H.S)
B)y^2 + yq2 – p^2 = 0
Sum = Q^2 = root1+root2
=> (AB)^2 ( Not equal to R.H.S)
C)y^2 – 2y + q^2 + p^2 = 0
Sum = root1+root2
=> 2 ( Not equal to R.H.S)
D)y^2 + 2qy – q^2 – p^2 = 0
Sum = 2Q = root1+root2
2AB (equal to SUM R.H.S)
Product = -Q^2-P^2
=> -AB^2-A^2-B^2-2AB (Not equal to Product R.H.S)
E)y^2 – 2qy + q^2 – p^2 = 0
Sum = 2Q = root1+root2
=> 2AB (equal to SUM R.H.S)
Product = +Q^2-P^2 = (Q+P)(Q-P)
=> (AB+A+B)(AB-A-B) (equal to Product R.H.S)
Answer is E
