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
142628281991919182726(35262627282
Answer:
x = 4/5
Step-by-step explanation:
Group like terms together. Start by subtracting 1 from both sides, obtaining
5x = -4 + 10x.
Then combine the x terms, obtaining 4 = 5x.
Solving for x by dividing both sides by 5, we get x = 4/5.
