Answer :

dashataran

Answer:

[tex]\text{D. }x=32\text{ and }x=0[/tex]

Step-by-step explanation:

[tex](x-16)^2=256\\\\\text{or, }(x-16)^2=16^2\\\\\text{or, }(x-16)^2-16^2=0\\\\\text{or, }(x-16-16)(x-16+16)=0\hspace{2cm}[a^2-b^2=(a-b)(a+b)]\\\\\text{or, }(x-32).x=0\\\\\text{i.e. }x=0\text{ or }32.[/tex]

Answer:

D

Step-by-step explanation:

given the quadratic equation

(x - 16)² = 256 ( take square root of both sides )

[tex]\sqrt{(x-16)^2}[/tex] = ± [tex]\sqrt{256}[/tex] , that is

x - 16 = ± 16 ( add 16 to both sides )

x = 16 ± 16

Then

x = 16 + 16 = 32

x = 16 - 16 = 0

The solutions are x = 32 and x = 0

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