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# Solve for k

3k3=27k

### 3 Answers by Expert Tutors

Patricia S. | Math Tutoring for K-12 & CollegeMath Tutoring for K-12 & College
5.0 5.0 (39 lesson ratings) (39)
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Hi, Lila!

Vivian and Ken are correct in that two of the roots are +/-3.  However, you need to be careful "canceling" out a k on both sides, as this removes one of the possible roots.  This type of problem is traditionally solved using factoring:

3k= 27k

Rearrange the equation so that it equals zero, just as you would do with a quadratic equation.
3k- 27k = 0

Factor out the LCM: 3k
3k(k- 9) = 0

Factor the quadratic expression in the parenthesis.
3k(k+3)(k-3) = 0

Set each part equal to 0 and solve for k.
3k = 0  --> k = 0 (this is the root that is removed when you "cancel" out a k early on)
k + 3 = 0  --> k = -3
k - 3 = 0  --> k = 3

This method gives you all 3 roots: k = -3, 0, and 3.

Patty

Hi Patricia;
I learned my lesson.  When establishing roots, do not cancel or eliminate anything.
Ken G. | MS Math Colgate Phi Beta Kappa- Tutor ALL Ages - SAT,ACT,SSAT,ASTBEMS Math Colgate Phi Beta Kappa- Tutor ...
4.6 4.6 (21 lesson ratings) (21)
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Divide both sides by K leaving
3k2=27
Divide both sides by 3...
(3k2)/3=27/3

Resulting in
k2=9

Square-root both sides...
k=+/-3
Vivian L. | Microsoft Word/Excel/Outlook, essay composition, math; I LOVE TO TEACHMicrosoft Word/Excel/Outlook, essay comp...
3.0 3.0 (1 lesson ratings) (1)
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Hi Lila;
3k3=27k
First, you have k3 on the left side, and k1 on the right side.  k1 cancels on both sides...
3k2=27
Divide both sides by 3...
(3k2)/3=27/3
k2=9
Square-root both sides...
k=+/-3