Answer:
no identity will escape! I know where you live
Answer:
2ab(3b^2+2a+4)
Step-by-step explanation:
6ab^3 + 4a^2b + 8ab
2*3*a*b*b^2 +2*2*a*a*b +2*2*2*a*b
Factor out the common terms
2ab( 3*b^2 +2*a +2*2)
2ab(3b^2+2a+4)
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
