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.
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Answer: x^2 -10x + 24
Step-by-step explanation:
(x-6) (x-4)
= x^2 - 6x - 4x + 24
= x^2 -10x + 24
The coordinates of the midpoint are (-6,3)
Answer:
The answer is 6 pounds.
Step-by-step explanation:
I figured it out by getting it wrong
Answer:
Equation in square form:
Extreme value:
Step-by-step explanation:
We are given
we can complete square
we can use formula
now, we can add and subtract 5^2
So, we get equation as
Extreme values:
we know that this parabola
and vertex of parabola always at extreme values
so, we can compare it with
where
vertex=(h,k)
now, we can compare and find h and k
we get
h=-5
k=-4
so, extreme value of this equation is
