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

[anemone]

Given that $a$ and $b$ are the roots of the equation $x^2+2000x+1=0$ whereas $c$ and $d$ are the roots of the equation $x^2-2008x+1=0$.

Evaluate $(a+c)(b+c)(a-d)(b-d)$.

 

[kaliprasad]

My attempt

Spoiler

We are given a,b are roots of equation so $f(x) = (x -a)(x-b) = x^2 + 2000x +1 = 0$
so $(a+c)(b+c) = (-c - a)(-c-b) = f(-c) = c^2 - 2000c + 1$

as c is root of $x^2-2008x + 1 = 0$ so $c^2 - 2000c +1 = 0$

so $(a+c)(b+c) = 0$ and multiplying by $(a-d)(b-d)$ we get $(a+c)(b+c)(a-d)(b-d) = 0$

 

[anemone]

Hi Kali...

Spoiler

Your answer is not correct...I am sorry.

 

[kaliprasad]

Spoiler

Thanks anemone., I mixed up 2000 and 2008. I have solved it incorrectly

 

[MarkFL]

Given that $a$ and $b$ are the roots of the equation $x^2+2000x+1=0$ whereas $c$ and $d$ are the roots of the equation $x^2-2008x+1=0$.

Evaluate $(a+c)(b+c)(a-d)(b-d)$.

Spoiler: My solution

Let's first look at the product:

\(\displaystyle (a+c)(b-d)=ab-ad+bc-cd\)

By Viete, we know:

\(\displaystyle ab=cd=1\)

Hence:

\(\displaystyle (a+c)(b-d)=bc-ad\)

Next, let's look at the product:

\(\displaystyle (b+c)(a-d)=ab-bd+ac-cd=ac-bd\)

And so we find:

\(\displaystyle (a+c)(b+c)(a-d)(b-d)=(bc-ad)(ac-bd)=abc^2-b^2cd-a^2cd+abd^2=c^2+d^2-(a^2+b^2)\)

Now, also by Viete, we have:

\(\displaystyle a+b=-2000\)

Hence:

\(\displaystyle a^2+2ab+b^2=2000^2\implies a^2+b^2=2000^2-2\)

Likewise:

\(\displaystyle c+d=2008\)

\(\displaystyle c^2+2cd+d^2=2008^2\implies c^2+d^2=2008^2-2\)

And so:

\(\displaystyle (a+c)(b+c)(a-d)(b-d)=(2008^2-2)-(2000^2-2)=2008^2-2000^2=8(4008)=32064\)

 

[anemone]

Very good job, MarkFL!

 

[anemone]

Spoiler: My solution

By Vieta's formula, we have $a+b=-2000; ab=1; cd=1$.

$\begin{align*}(a+c)(b+c)(a-d)(b-d)&=[ab+(a+b)c+c^2][ab-(a+b)d+d^2]\\&=(1-2000c+c^2)(1+2000d+d^2)\\&=(c^2-2008c+1+8c)(d^2-2008d+1+4008d)\\&=8c(4008d)\\&=32064\end{align*}$