Respuesta :

wegnerkolmp2741o

Answer:

one real solution

Step-by-step explanation:

We can use the discriminant to find the types of solutions

b^2 -4ac  where    ax^2 +bx+c=0

Since

x^2 -8x+16 =0

we know a=1 b=-8 and c=16

(-8)^2 - 4(1)(16)

64 -64 = 0

If the discriminant > 0 we have two real solutions

If the discriminant = 0 we have 1 real solutions

If the discriminant < 0 we have two complex solutions

Since it equals zero, we have one real solution

S1NGH

Answer:

One real solution because the answer is [tex](x-4)^{2}[/tex]

Step-by-step explanation:

[tex]x^{2} -8x+16=\\(x-4)(x-4)[/tex]

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