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.
<h3>What is the condition for a solution?</h3>
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.
To learn more about the solution visit:
brainly.com/question/1397278
It is like if you have two number lines. Then one has a pointer at 0, and the other one has the pointer facing at 2 then it will be lesser because the first one has 0 or if it turned around it would be greater. If it is equal then it would have the same number, but watch out if it is a negative number.
Answer:
it is the answer. let me do research and I will get answer
Its -2.5875
hoped it helped
