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.

$->9-40k >0$

$:.40k<9$

$k<9/40$

How do you find all values of k such that 2x^2-3x+5k=02x^2-3x+5k=0 has two solutions?