Respuesta :

EjerciciosFyQ
[tex]x^2 - 8x + 16 = 0\ \to\ a = 1\ ;\ b = -8\ ;\ c = 16[/tex]

[tex]\frac{-b\pm \sqrt{b^2 - 4ac}}{2a} = \frac{8\pm \sqrt{64 - 4\cdot 1\cdot 16}}{2\cdot 1} = x_1 = 4\ ;\ x_2 = 4[/tex]

[tex]x^2 - 8x + 16 = (x - 4)^2[/tex]
dalendrk
[tex]x^2-8x+16=0\\\\x^2-2x\cdot4+4^2=0\ \ \ /use\ (a-b)^2=a^2-2ab+b^2/\\\\(x-4)^2=0\iff x-4=0\to\boxed{x=4}[/tex]

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