How do you find the discriminant and how many and what type of solutions does #x^2-8x+16=0# have?

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Answer 1 · 2015-05-03T07:36:50

The equation is of the form #color(blue)(ax^2+bx+c=0# where:

#a=1, b=-8, c=16#

The Disciminant is given by :
#Delta=b^2-4*a*c#
# = (-8)^2-(4*1*16)#
# = 64-64=0#

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 = 0#, this equation has ONE REAL SOLUTION

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

As #Delta = 0#, #x = -b/(2a) = -(-8)/(2*1) = 8/2 = 4#

#x = 4# is the solution