SOLUTION: If 6k^2 + k = 2 and k > 0, find the value of k.

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: If 6k^2 + k = 2 and k > 0, find the value of k.       Log On


   



Question 1208922: If 6k^2 + k = 2 and k > 0, find the value of k.
Found 3 solutions by mccravyedwin, ikleyn, math_tutor2020:
Answer by mccravyedwin(407) About Me  (Show Source):
You can put this solution on YOUR website!

Here's one just like it:

If 12k^2 + k = 6 and k > 0, find the value of k.

12k%5E2%2Bk=6
12k%5E2%2Bk-6=0
Factor:
%284k%2B3%29%283k-2%29=0
4k+3=0; 3x-2=0
4k=-3;    3k=2 
 k=-3/4    k=2/3

2/3 is the answer to this problem.
Now do yours the same way.

Edwin




Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
.

Write this equation in the standard quadratic form

    6k^2 + k - 2 = 0.


To find k, apply the quadratic formula

    k%5B1%2C2%5D = %28-1+%2B-+sqrt%28%28-1%29%5E2+-+4%2A6%2A%28-2%29%29%29%2F%282%2A6%29 = %28-1+%2B-+sqrt%2849%29%29%2F12 = %28-1+%2B-+7%29%2F12.


They want k  be positive - so take this unique positive root  k = %28-1%2B7%29%2F12 = 6%2F12 = 1%2F2.


At this point, the solution is complete.


ANSWER.  k = 1%2F2.

Solved.

------------------------

On solving quadratic equations using the quadratic formula,  see the lessons
    - Introduction into Quadratic Equations
    - PROOF of quadratic formula by completing the square
in this site.


If factoring requires mental efforts from you more than 5 - 7 - 10 seconds
and is not requested in the assignment, use the quadratic formula.

It is the general rule to save your mind from unnecessary exercises for real work.



Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

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: k = 1/2