Difference of 2 roots

Question

$2003^2x^2 - (2002)(2004)x - 1 = 0$

$x^2+2002x - 2003 = 0$

What is the difference of the 1st equation's larger root and the smaller root of the 2nd equation?

I did not know how to approach this, so I used the quadratic formula but could not easily simplify the expressions. What properties will be used ?

Answer 1

One solution of the $1$. equation is $x_1=1$ and by Viete we have $$x_1+x_2 ={2003^2-1\over 2003^2}= 1-{1\over 2003^2}$$

so $x_2 = {1\over 2003^2} <1=x_1$

For the second equation we have also $x'_1=1$ and again by Viete we have $x_2' =-2003$ so...

Answer 2

\begin{align*} 2003^2x^2-2002(2004)(x)-1&=2003^2x^2-(2003-1)(2003+1)x-1\\ &=2003^2(x^2-x)+x-1\\&=(x-1)(2003^2x-1) \end{align*}

\begin{align*} x^2+2002x-2003=(x-1)(x+2003) \end{align*}