How do you find the discriminant, describe the number and type of root, and find the exact solution using the quadratic formula given #x^2-x+6=0#?

1 Answer
George C.
Jan 5, 2017

#x = 1/2+-sqrt(23)i#

Explanation:

The discriminant #Delta# of a quadratic equation with rational coefficients gives us the following information:

  • If #Delta > 0# then there are two, distinct, real roots. If #Delta# is a perfect square, then they are both rational.

  • If #Delta = 0# then there is one, repeated, rational, real root.

  • If #Delta < 0# then there are two, distinct, non-Real, Complex roots, which form a complex conjugate pair.

The quadratic equation:

#x^2-x+6=0#

is in the form:

#ax^2+bx+c=0#

with #a = 1#, #b=-1# and #c=6#

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = (color(blue)(-1))^2-4(color(blue)(1))(color(blue)(6)) = 1-24 = -23#

Since #Delta < 0# this quadratic has no Real roots. It has a complex conjugate pair of non-Real, Complex roots.

We can find the roots using the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#color(white)(x) = (-b+-sqrt(Delta))/(2a)#

#color(white)(x) = (1+-sqrt(-23))/(2*1)#

#color(white)(x) = (1+-sqrt(23)i)/2#

#color(white)(x) = 1/2+-sqrt(23)i#

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