Let α be the common root of the equation 3x2−7x+2=0 and kx2+7x−3=0, so 3α2−7α+2=0 ...(i) and ‌‌kα2+7α−3=0....(ii) By cross multiplication, we get ‌
α2
21−14
=‌
−α
−9−2k
=‌
1
21+7k
⇒‌‌‌
α2
7
=‌
α
2k+9
=‌
1
7(k+3)
∴‌‌‌
1
k+3
=‌
(2k+9)2
49(k+3)2
∵k is positive, so 49(k+3)=4k2+36k+81 ⇒‌‌4k2−13k−66=0 ⇒‌‌4k2−24k+11k−66=0 ⇒‌‌4k(k−6)+11(k−6)=0 ⇒‌‌k=6‌‌[∵k is positive ]