Respuesta :

kate200468
[tex]x^2 + 16 = 8x\\\\x^2-8x+16=0\\\\x^2-4x-4x+16=0\\\\x(x-4)-4(x-4)=0\\\\(x-4)(x-4)=0\\\\(x-4)^2=0\ \ \ \Leftrightarrow\ \ \ x-4=0\ \ \ \Leftrightarrow\ \ \ x=4\\\\ Ans.\ One\ solution[/tex]
AL2006
Every equation that has an ' x² ' in it has two solutions.

If it's a perfect square, then the solutions are equal, and they look like only 1.

Both solutions to this equation are [ x = 4 ] .

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