?

Given the quadratic equation kx² + 2x + 4 = 0, find the value of k so that the roots are equal.

How do I do this?

The "quadratic equation" to solve ax^2 + bx + c = 0:
x = (-b +/- sqrt(b^2 - 4*a*c*))/2*a

gives two values of x exactly when
(b^2 - 4*a*c*) > 0

one value of x exactly when
(b^2 - 4*a*c*) = 0

and no (real) values of x exactly when
(b^2 - 4*a*c*) < 0

hence you want (b^2 - 4*a*c*) = 0
substitute in b=2, a=k, c=4 and you're away.

>>2
Thanks!

For what values of m does mx^2 - 2x - 4 = 0 have real and distinct roots?

How would this one be done?

That's when (b^2 - 4*a*c*) > 0
because that expression is square rooted when solving for roots, and you'd get imaginary bits if it was < 0 (square root of negative).  The distinctness comes from one solution uses plus that expression, and the other solution uses minus that expression, yielding two different roots.