Question

If k is a natural number and the roots of the equation x^2 + 11x+ 6k= 0 are rational numbers, then find the value of k

Answer 1

$x {}^{2} - 11x + 6k = 0$
Comparing eq with
$ax { }^{2} + bx + c = 0$
a=1;b=-11;c=6k
Roots of the given equations are rational.
$so \: b {}^{2} - 4ac \: isa \: perfect \: square$
so(-11)^2-4×1×6k is a perfect square
121-24k is a perfect square.
121-24=97 is not perfect square so k not equal to 1.
121-48=73 is not perfect square so k not equal to 2.
121-72=49 is a perfect square so k can be 3.
121-96=25 is a perfect square so k can be 4.
121-120=1 is a perfect square so k can be 5.
k cannot be more than 5 as b^2-4ac>=0
So,k={3,4,5}