NIMCET 2008 Question Paper with solutions

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Question : 7 of 120

Marks: +1, -0

If a, b are the roots of

x^{2}+p x+1=0

and

c, d

are roots of

x^{2}+q x+1=0

the value of

E=(a-c)(b-c)(a+d)(b+d)

is

$p^{2}-q^{2}$
$q^{2}-p^{2}$
$q^{2}+p^{2}$
None of these

Solution:

a. b are roots of

x^{2}+p x+1=0

⇒ a+b=-p ; a b=1

...(i)
Also, c and d are roots of

x^{2}+q x+1=0

⇒ c+d=-q ; c d=1

.....(ii)
Now,

E=(a-c)(b-a)(a+d)(b+d)

=(a-c)(b+d)(b-c)(a+c)

=(a b-c d-b c+a d)(a b-a c+b d-c d)( ∵ a b=c d)

=(a d-b c)(b d-a c)

=a b(d^{2}+c^{2})-c d(a^{2}+b^{2})

=a b\{(c+d)^{2}-2 c d\}-c d\{(a+b)^{2}-2 a b\}

=(q^{2}-2)-(p^{2}-2)

=q^{2}-p^{2}

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