If one root of the equation ax2 + bx + c = 0 be n times the other root, then
na2 =bc (n + 1)2
nb2 = ac (n+1)2
nc2 = ab (n+1)2
None of these
Let the roots are α and nα
Sum of the roots, α + nα = - ba ⇒α = - ba(n + 1) ..................... (i) and product, α.nα = ca ⇒α2 = cna .................. (ii) From (i) and (ii), we get ⇒- ba(n+1)2 = cna ⇒b2a2 (n +1)2= cna ⇒nb2 = ac (n + 1)2