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Solve for k

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3 Answers

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

Comments

Hi Patricia;
I learned my lesson.  When establishing roots, do not cancel or eliminate anything.
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