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
What is the volume of the composite figure? (Use for .) Do NOT round your answer
answer- 2388.16
in- 3
Answer:
after what.................. ???
For there to be an infinite number of solutions, the quantity on the left side of the equation must be the same as on the right.
First, distribute the equation to get
6x + 18 = 3xh + 9h
If h = 2, the equation on the right would also be 6x + 18 which would yield the same equation and hence an infinite number of solutions
So the answer is h = 2
