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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brainly.com/question/1397278
Answer:
y = ![\frac{1}{2}x + 3](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dx%20%2B%203)
Step-by-step explanation:
The equation for a linear graph is usually written in the following format...
y = mx + b
Where m would be the slope, usually referring to rise over run and b would be the y-intercept (where the line crosses the y-axis). From the graph, we can see that the line crosses the y-axis at point 3 so b would be 3. The graph also shows us that for every 1 value that the line rises it moves to the right 2 values. Therefore, the slope would be 1/2. Using these values we can create the following equation...
y = ![\frac{1}{2}x + 3](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dx%20%2B%203)
No 2/3 of $21 is $14 so 6x14 is $84 less than $110
The answer is 11:5 ……………….
Replace x and y with the given values and solve the equations:
A) 3(5) + 2 = 17
B) 2(5)^2 = 50
C) (5-2)^2 = 9