Share:
Notifications
Clear all

[Solved] Integers quadratic equation

1 Posts
1 Users
0 Reactions
487 Views
0
Topic starter

Find positive integers  m,n such that the quadratic equation 4x^2-2mx+n=0  has two real roots both of which are between 0 and 1.

1 Answer
0
Topic starter

The equation has two real roots, thus \small \Delta =4m^2-16n\geq 0\Rightarrow n\leq m^2/4.

Since both roots are between 0 and 1, then \small f(0)=n> 0, f(1)=4m-2m+n>0,  then  \small n>2m-4 (m,n \epsilon \mathbb{N})

Hence, \small 2m-4<n\leq m^2/4,  which implies a unique choice: 

\small m=2, n=1.

Share: