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If α,β are the zeros of x2 + px + 1 and γ,δ be those of x2 + qx + 1, then the value of (α–γ) (β–γ) (α+δ) (β+δ) = [DCE-2000]
- a)p2 – q2
- b)q2 – p2
- c)p2
- d)q2
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
If α,βare the zeros of x2 + px + 1 and γ,δbe th...
Alpha(a) and beta(b) are roots of x^2 + px + 1
This implies that sum of roots= a+b = -p/1=-p
And the product of roots = ab = 1/1=1
Similarly ,
Gamma(c) and delta(d) are roots of x^2 + qx + 1
So c+d=-q and cd =1.
The above results can be obtained once we know that any quadratic equation has two roots and hence can be written as (x-p)(x-q)=0 where a and b are the roots .
So x^2 -(p+q)x + pq =0
Comparing this with the general form of quadratic equation :ax^2 + bx + c= 0 we get
Sum of the roots =p+q= -b/a
And product of the roots = pq = c/a}
RHS=(a-c)(b-c)(a+d)(b+d)
=(c^2-ac-bc +ab)(d^2 +bd +ad + ab)
We know ab=1
So RHS= (c^2-ac-bc +1)(d^2 +bd +ad + 1)
= (c^2)(d^2) +(a+b)c^2(d) + c^2 -(d^2)c(a+b) -(a+b)^2(cd) -(a+b)c + d^2 + bd + ad + 1
= 1 + ac+bc + c^2 - da-db - (a^2 + b^2 + 2(1)) -ac -bc + bd + ad +1
Cancelling off all the common terms,
We get c^2 +d^2-a^2-b^2
= c^2+d^2-a^2-b^2
=c^2 -a^2 + d^2 -b^2 + 2(1) -2(1){ ab=cd =1}
=c^2 + d^2 + 2cd - a^2 - b^2 - 2ab
LHS=(c+d)^2-(a+b)^2
Therefore,
But we know -p= a+b and -q = c+d
LHS= q^2-p^2
= c^2+d^2-a^2-b^2
=c^2 -a^2 + d^2 -b^2 + 2(1) -2(1){ ab=cd =1}
=c^2 + d^2 + 2cd - a^2 - b^2 - 2ab
LHS=(c+d)^2-(a+b)^2
Therefore,
But we know -p= a+b and -q = c+d
LHS= q^2-p^2
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If α,βare the zeros of x2 + px + 1 and γ,δbe those of x2 + qx + 1, then the value of (α–γ) (β–γ) (α+δ) (β+δ) = [DCE-2000]a)p2– q2b)q2– p2c)p2d)q2Correct answer is option 'B'. Can you explain this answer?
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