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

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