Let α be the common root of both the given equations. Then a satisfies both the equations. So,
α2 – kα – 21 = 0 ...(i)
a2 – 3kα + 35 = 0 ...(ii)
Solving equations (i) and (ii) simultaneously, we get
\(\frac{α^2}{-35k-63k}=\frac{α}{-21-35}=\frac{1}{-3k+k}\)
⇒ α2 = \(\frac{-98k}{-2k}\) = 49 and α = \(\frac{-56k}{-2k}\) = \(\frac{28}{k}\)
∴ 49 = \(\big(\frac{28}{k}\big)^2\) ⇒ k2 = \(\frac{28\times28}{49}\) = 16 ⇒ k = ± 4.