Let the roots be r and 6r. Substituting each root in the given quadratic equation gives
k r^{2} − 14r + 8 = 0 (Eq. 1) and
36k r^{2} − 84r + 8 = 0 (Eq. 2).
Multiplying the first equation above by 36 throughout gives
36k r^{2} − 504r + 288 = 0 (Eq. 3).
On subtracting Eq. 3 from Eq. 2, we get
420r − 280 = 0 or r = 2/3.
So, the two roots are 2/3 and 4.
Substituting x = 4 in Eq. 1 gives
16k − 56 + 8 = 0 or k = 3.
bsl 258 days ago
