Question 111499
{{{(x-4)^2=0}}}
{{{x^2 - 8x + 16 = 0}}}
this is in the form {{{ax^2 + bx + c = 0}}}
a = 1
b = -8
c = 16
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
As long as {{{b^2 - 4ac}}} is a positive number, the roots are real
If {{{b^2 = 4ac}}}, there is a single real root
{{{b^2 - 4ac = (-8)^2 - 4*1*16}}}
{{{b^2 - 4ac = 64 - 64}}}
{{{b^2 - 4ac = 0}}}
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{x = (-(-8) +- sqrt(0 ))/(2*1) }}}
{{{x = 8/2}}}
{{{x = 4}}} answer