Grade 11Algebra
Let the roots be α and n α
Sum of roots, α + nα = -ba ⇒ α = -ba(n+1) ......(i)
and product, α.n.α = 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.
Let the roots be α and n α
Sum of roots, α + nα = -ba ⇒ α = -ba(n+1) ......(i)
and product, α.n.α = 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.
