Question

How do you use the discriminant to determine the numbers of solutions of the quadratic equation `2x^2-6x+5 = 0` and whether the solutions are real or complex?

Answer 1

`Delta = -4 < 0`, so this quadratic has a pair of non-Real Complex solutions.

Explanation:

`2x^2-6x+5=0`

is in the form:

`ax^2+bx+c = 0`

with `a=2`, `b=-6` and `c=5`

This has discriminant `Delta` given by the formula:

`Delta = b^2-4ac = (-6)^2-4(2)(5) = 36-40 = -4`

Since `Delta < 0`, this quadratic equation has no Real solutions. It has a complex conjugate pair of distinct non-Real Complex solutions.

We can find the solutions by completing the square:

`0 = 2(2x^2-6x+5)`

`color(white)(0) = 4x^2-12x+10`

`color(white)(0) = 4x^2-12x+9+1`

`color(white)(0) = (2x-3)^2-i^2`

`color(white)(0) = ((2x-3)-i)((2x-3)+i)`

`color(white)(0) = (2x-3-i)(2x-3+i)`

Hence solutions:

`x = 3/2+1/2i" "` and `" "x = 3/2-1/2i`