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
Answer:
110 students in the class
Step-by-step explanation: If this is helpful and correct please mark as brainliest!
3\4
because the spaces are bigger draw a model and you will see
Answer:
It increases
Step-by-step explanation:
(5/x) + 5
As x decreases, 5/x increases. So the expression increases.
Hey there!
Let's break this expression into two parts:
6(3x-1) and -10x
To solve the first part, we need to use the distributive property which states:
a(b+c) = ab+ac
Applying that to this problem, we have:
6(3x) + 6(-1) =
18x - 6
Now, we can take that -10x and put it right back in:
18x - 6 - 10x
Combine like terms and subtract the 10x from the 18x to get:
8x - 6
Hope this helps!
