Question

How to use the discriminant to find out how many real number roots an equation has for `4/3x^2 - 2x + 3/4 = 0`?

Answer 1

`(4/3)x^2-2x+3/4` is of the form `ax^2+bx+c`

with `a=4/3`, `b=-2` and `c=3/4`.

The discriminant `Delta` is given by the formula `b^2-4ac`

In our case:

`Delta = b^2-4ac`

`= (-2)^2 - 4*(4/3)*(3/4)`

`= 4 - 4 = 0`

If `Delta > 0` then our equation would have 2 real roots.

If `Delta < 0` then our equation would have 2 complex non-real roots.

In our case `Delta = 0` implies that our equation has 1 repeated real root.

That root is

`x = (-b +- sqrt(Delta)) / (2a) = (-(-2)+-0)/(2*(4/3)) = 2/(2*(4/3)) = 1/(4/3) = 3/4`