3p2x3+px2+qx+3=0 p=1 and q=−7 ⇒‌‌3x3+x2−7x+3=0.....(i) On putting x=1 3×13+12−7×1+3=7−7=0 ∴(x−1) is one factor in the Eq. (i),     3x2+4x−3     ────────── (x−1)3x3+x2−7x+3     +3x3−3x2      −‌‌‌‌‌‌+    ────────────     4x2−7x+3     4x2∓4x     −‌‌‌+    ────────────       −3x+3       −3x+3       +‌‌‌‌−    ─────────────        0 ∴ Eq. (i) can be written as (x−1)(3x2+4x−3)=0 3x2+4x−3=0 gives the irrational roots. ⇒α+β‌=‌