Question

Find all values of k for which the equation 3x^2−4x+k=0 has no solutions.

Answer 1

The given equation has no solution when K is any real number  and k>12

We have given that

3x^2−4x+k=0

△=b^2−4ac=k^2−4(3)(12)=k^2−144.

What is the condition for a solution?

If Δ=0, it has 1 real solution,

Δ<0 it has no real solution,

Δ>0 it has 2 real solutions.

We get,

Δ=k^2−144 here Δ is not zero.

It is either >0 or <0

Δ<0 it has no real solution,

Therefore the given equation has no solution when K is any real number.

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Answer 2

Answer:

k>1 1/3

Step-by-step explanation:

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