Question 1208922
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6k^2+k = 2
6k^2+k-2 = 0


You can factor or use the quadratic formula as the other tutors have shown.


Another approach is to use the Rational Root Theorem.
This is where we divide factors of q over factors of p
p = 6 = leading coefficient
q = -2 = last term of the polynomial


Factors of 6 are: 1,2,3,6
Factors of 2 are: 1,2


Dividing those terms gives these possible rational roots
1, 1/2, 1/3, 1/6, 2, 2/3
-1, -1/2, -1/3, -1/6, -2, -2/3
We consider the plus minus of each.


Then plug each item into 6k^2+k-2 to see if we get zero or not.
Try k = 1 to get 6k^2+k-2 = 6*(1)^2+1-2 = 5
The nonzero result means that k = 1 is not a root.


Now let's try k = 1/2.
6k^2+k-2 = 6*(1/2)^2+(1/2)-2 = 0
We get 0 as the result, so it proves k = 1/2 is a root.
The other root is k = -2/3
All of the other values will lead to nonzero results, so we can eliminate them. Also keep in mind that any quadratic will have 2 roots.


We have shown the two roots are k = 1/2 and k = -2/3
Ignore k = -2/3 since k > 0 was mentioned in the instructions.


Answer: <font color=red>k = 1/2</font>
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