If α≠0 and 0 are the roots of x2−5kx+(6k2−2k)=0, then find α. Step 1: Use the fact that 0 is a root. If x=0 is a root, it should satisfy the equation: ‌02−5k(0)+(6k2−2k)=0 ‌⇒6k2−2k=0 ‌⇒2k(3k−1)=0 So k=0 or k=‌
1
3
. But if k=0, the equation becomes x2=0, giving a double root 0 - contradicting that one root is nonzero (α≠0). Hence, k=‌
1
3
. Step 2: Substitute k=‌
1
3
into the equation. ‌x2−5(‌
1
3
)x+(6(‌
1
3
)2−2(‌
1
3
))=0 ‌⇒x2−‌
5
3
x+(‌
6
9
−‌
2
3
)=0 ‌⇒x2−‌
5
3
x+(‌
2
3
−‌
2
3
)=0 Wait - compute the constant term carefully: ‌
6
9
−‌
2
3
=‌
2
3
−‌
2
3
=0. So the equation becomes: ‌x2−‌
5
3
x=0 ‌x(x−‌
5
3
)=0 Step 3: Identify the roots. Roots are x=0 and x=‌
5
3
. Given that one root is 0 and another is α≠0, α=‌
