How do you find all values of k such that #2x^2-3x+5k=0# has two solutions?

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Answer 1 · 2017-02-16T01:12:33

For #x in CC: k!=9/40#

For #x in RR: k<9/40#

#2x^2-3x+5k = 0#

The discriminant here is: #(-3)^2 - 4xx2xx5k#

#=9-40k#

Assume #k in QQ#

For the equation to have two roots (real and complex) the discriminant must not equal zero.

#->9-40k !=0#

#k!=9/40#

If the roots are to be real only the discriminant must be greater than zero.

#->9-40k >0#

#:.40k<9#

#k<9/40#