What must the values of k bear so that :

Question

2x^3+(k_3)x+k_3=0 has one real root?two real root? no real root?

Answer 1

I suspect you meant 2x^^2+(k-3)x+k-3=0, since cubics always have at least one real root.

for one real root, the discriminant must be zero, so
(k-3)^2 -8(k-3) = 0
(k-3)(k-3-8) = 0
so
k-3=0 ==> k=3
or
k-3 = 8 ==> x=11
check:
k=3: 2x^2 = 0 ✅
k=11: 2x^2+8x+8 = 2(x+4)^2 = 0 ✅

Now do the other two cases. Post your work if you get stuck.