How many solutions does the equation #x^2 - 16x + 64 = 0# have?

2 Answers
Aug 19, 2016

Answer:

Conventionally, we say that the eqn. has #2# identical roots : #8&8#.

Explanation:

#x^2-16x+64+0 rArr (x-8)^2=0#

#rArr x=8, &, x=8#

Conventionally, we say that the eqn. has #2# identical roots : #x=8,8#.

Aug 20, 2016

For the quadratic equation #ax^2+bx+c=0#, the discriminant #Delta=b^2-4ac# tells us about the nature of the equation's roots:

  • If #Delta>0#, the equation has two real roots.
  • If #Delta=0#, the equation has a single double root.
  • If #Delta<0#, the equation has no real solutions.

So, for #x^2-16x+64=0#, we see that #a=1#, #b=-16#, and #c=64#. Thus:

#Delta=b^2-4ac=(-16)^2-4(1)(64)=256-256=0#

Since #Delta=0#, the equation has one single solution.