Question

If roots of 2xsquare-6x+k=0 are real and equal find k

Answer 1

this is example
How do you find all values of k such that 2x2−3x+5k=0has two solutions?



Answer:

For x∈C:k≠940

For x∈R:k<940

Explanation:

2x2−3x+5k=0

The discriminant here is: (−3)2−4×2×5k

=9−40k

Assume k∈Q

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

→9−40k≠0

k≠940

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

→9−40k>0

∴40k<9

k<940

I hope this will help you