Determine the values of K for the equation

Question

Determine the values of K for which 3x²+kx+12=0 has each type of solution justify each answer.

a) exactly one real solution

b)two imaginary solutions

c)two real solutions

Answer 1

Nina,

 

This is a quadratic equation whose solution is given by the quadratic formula, where A=3, B=k, and C=12.

The formula gives the solutions as:

 

x = {-B ± √(B² - 4AC)} /2A

 

It is the square root portion that determines how many and what kind of distinct solutions you will get.

So  (B² - 4AC) = [k² - 4(3)(12)] = [k² - 144] 

 

(a) For one real solution, both solutions must be identical, so the square root must be zero, and 

     [k² - 144] = 0.  Hence k = ± 12.  And x = -B/2A = -k/6.  

     When k = +12, x₁=x₂ = -2 and when k = -12 then x₁=x₂ = +2.

 

(b) For two imaginary solutions, you must have a negative non-zero real number inside the square root.

     So  [k² - 144] < 0  and -12 < k < 12.  Then x = {-k ± i √(144 - k²)} /6  since bringing the imaginary i outside the

     square root makes the square root positive given the limits on k.  (Recall that √-2 = √(-1)(+2) = i √+2).

 

(c) For two real solutions, you must have a positive non-zero real number inside the square root.  This will happen when

     |k| > 12.  Then x = {-k ± √(k² - 144)} /6.

 

BTW, remember that a quadratic polynomial has at most 2 distinct solutions.