How do you solve #x^2 - 12x + 52 = 0#?

2 Answers
Jun 8, 2015

#x^2 - 12x + 52 = 0#

The equation is of the form #color(blue)(ax^2+bx+c=0# where:
#a=1, b=-12, c=52#

The Discriminant is given by:
#Delta=b^2-4*a*c#
# = (-12)^2-(4*1*52)#
# = 144 - 208#
# = -64#

If #Delta=0# then there is only one solution.
(for #Delta>0# there are two solutions,
for #Delta<0# there are no real solutions)

As #Delta = -64#, this equation has NO REAL SOLUTIONS

Note :
The solutions are normally found using the formula
#x=(-b+-sqrtDelta)/(2*a)#

As #Delta = -64#, #x = (-(-12)+-sqrt(-64))/(2*1) = (12+-sqrt(-64))/2#

Jun 8, 2015

#x^2-12x+52# is of the form #ax^2+bx+c# with #a=1#, #b=-12# and #c=52#. This has discriminant given by the formula:

#Delta = b^2-4ac = (-12)^2-(4xx1xx52)#

#=144-208=-64 < 0#

So #x^2-12x+52=0# has no real roots. It has two distinct complex roots.