An algebra problem by Wildan Bagus Wicaksono

Algebra Level 3

Let α \alpha be the larger root of ( 2004 x ) 2 − 2003 ⋅ 2005 x − 1 = 0 (2004x)^2 - 2003 \cdot 2005x - 1 = 0 and β \beta be the smaller root of x 2 + 2003 x − 2004 = 0 x^2 + 2003x - 2004 = 0 . Determine the value of α − β \alpha -\beta .


The answer is 2005.

Wildan Bagus Wicaksono by Wildan Bagus Wicaksono , 18, Indonesia

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

( 2004 x ) 2 − 2003 ⋅ 2005 x − 1 = 0 200 4 2 x 2 − ( 2004 − 1 ) ( 2004 + 1 ) x − 1 = 0 200 4 2 x 2 − ( 200 4 2 − 1 ) x − 1 = 0 200 4 2 x 2 − 200 4 2 x + x − 1 = 0 200 4 2 x ( x − 1 ) + x − 1 = 0 ( 200 4 2 x + 1 ) ( x − 1 ) = 0 ⟹ α = 1 the larger root. \begin{aligned} (2004x)^2 - 2003 \cdot 2005 x - 1 & = 0 \\ 2004^2x^2 - (2004-1)(2004+1) x - 1 & = 0 \\ 2004^2x^2 - (2004^2-1)x - 1 & = 0 \\ 2004^2x^2 - 2004^2x + x - 1 & = 0 \\ 2004^2x(x-1) + x - 1 & = 0 \\ (2004^2x+1)(x-1) & = 0 \\ \implies \color{#3D99F6} \alpha & = 1 & \small \color{#3D99F6} \text{the larger root.} \end{aligned}

x 2 + 2003 x − 2004 = 0 ( x + 2004 ) ( x − 1 ) = 0 ⟹ β = − 2004 the smaller root. \begin{aligned} x^2+2003x-2004 & = 0 \\ (x+2004)(x-1) & = 0 \\ \implies \color{#3D99F6} \beta & = -2004 & \small \color{#3D99F6} \text{the smaller root.} \end{aligned}

⟹ α − β = 1 − ( − 2004 ) = 2005 \implies \alpha - \beta = 1 -(- 2004) = \boxed{2005}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...