The two roots of an equation x3−9x2+14x+24=0  are in the ratio 3:2, The roots will be

The given equation is x3−9x2+14x+24=0 Let the  required roots are 3α,2α,β         (∵  ratio of two roots are 3:2)∴ ∑α=3α+2α+β=−(−9)1=9 ⇒ 5α+β=9                            …….(i)   ∑αβ=3α.2α+2α.β+β.3α=14 ⇒ 5αβ+6α2=14                      ……(ii)and    αβγ=3α.2α.β=−24 ⇒ 6a2β=−24  or α2β=−4       .….(iii)From Eq. (i), β=9−5α , put the value of β  in Eq. (ii)⇒ 5α(9−5α)+6α2=14 ⇒ 19α2−45α+14=0 ⇒ (α−2)(19α−7)=0 ∴  α=2  or 719 ∴  from Eq. (i), if α=2 , then β=9−5×2=−1 ∵α=2,β=−1  satisfy the equation (iii) so required roots are 6,4,−1 .