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?

1 Answer
Dec 23, 2016

Answer:

#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#