Question

If a, b are the roots of x2+x+1=0 & c,d are the roots of x2+3x+1=0 then what is the value of (-c)(b+d)(a+d)(b-c) ?

Answer 1

Answer:

Step-by-step explanation:

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.

{side note:

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}

Now coming to the question

LHS= q^2-p^2

But we know -p= a+b and -q = c+d

Therefore,

LHS=(c+d)^2-(a+b)^2

=c^2 + d^2 + 2cd - a^2 - b^2 - 2ab

=c^2 -a^2 + d^2 -b^2 + 2(1) -2(1){ ab=cd =1}

= c^2+d^2-a^2-b^2

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= LHS.