If the equations 2x2+kx−5=0 and x2−3x−4=0 have one root in common, then k =

Let a be the common root of the two equationsThen,2α2+kα−5=0 and, α2−3α−4=0Solving these two equations, we getα2−4k−15=α−5+8=1−6−k⇒ α2=4k+15k+6 and α=−3k+6⇒ −3k+62=4k+15k+6 ∵α2=(α)2⇒ (4k+15)(k+6)=9⇒ 4k2+39k+81=0⇒k=−3 or ,k=−274