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
<h2>Answer:</h2><h2>let number be x</h2><h2>Step-by-step explanation:</h2><h2>5x - 1 = 24</h2><h2>5x = 25</h2><h2>x = 5</h2>
Answer:
x= - square root 141 or x= square root 141
It is not differentiable at x=1 since the slope of the tangent line as x -> 1 from the right is 1 while the slope of the tangent line as x->1 from the left is -1
Answer:
S is -2,-3 r -3,-4 R 0,-5
Step-by-step explanation:
is the reflection points :)
