Problem
ALG-B1-M03-P021 Parameter and an Integer Root
#21
★★★★☆ Level 4 of 5
Find all integers \(a\) for which \(x^2-(a+1)x+a+6=0\) has root \(x=3\) or \(x=4\).
Substitute \(x=3\) and \(x=4\) separately.
For \(x=3\): \(9-3(a+1)+a+6=0\), so \(12-2a=0\), hence \(a=6\).
For \(x=4\): \(16-4(a+1)+a+6=0\), so \(18-3a=0\), hence \(a=6\). Thus the only value is \(a=6\). For this value, both \(3\) and \(4\) are indeed roots.
A small problem on careful checking of conditions.