Find the values of K for which the equation 2x^2+3kx+9=0 has: a) No solution b)one solution c)two solutions?

Can you please show how you worked it out or explain it a little bit.

Find the values of K for which the equation 2x^2+3kx+9=0 has: a) No solution b)one solution c)two solutions?
Since the discriminant of the quadratic formula determines the Nature of roots. we can use it to solve for the value of k


If b² - 4ac %26gt; 0 two solutions


If b² - 4ac = 0 one solution


If b² - 4ac %26lt; 0 no solution





a = 2 b = 3k c = 9


(3k)² - 4(2)(9) = 0


9k² - 72 = 0


9k² = 72


k² = 8


k = 2√2


when k = 2√2 it has only one solution





k %26lt; -2√2 or k %26gt; 2√2 ---------------- 2 solutions


-2√2 %26lt; k %26lt; 2√2 --------- no solutions
Reply:check determinant


9k^2-72=0 for 1 soln


k=+-sqrt8


9k^2-72%26gt;0 for 2 soln


k^2%26gt;8


k%26gt;sqrt8 or k%26lt;-sqrt8





for no soln, -sqrt8%26lt;k%26lt;sqrt8