Solve for k

Question

3k³=27k 

Answer 1

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:

 

3k³ = 27k

 

Rearrange the equation so that it equals zero, just as you would do with a quadratic equation.

3k³ - 27k = 0

 

Factor out the LCM: 3k

3k(k² - 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

Answer 2

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

Answer 3

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