1)
[tex](x - 4) {}^{2} - 28 = 8 \\ (x - 4) {}^{2} = 8 + 28 \\ (x - 4) {}^{2} = 36[/tex]
This must be true for some value of x, since we have a quantity squared yielding a positive number, and since the equation is of second degree,there must exist 2 real roots.
[tex] \sqrt{(x - 4) {}^{2} } = ± \sqrt{36} \\x - 4 = ±6 \\ x _{1}- 4 = 6 \: \: \: \: \: \: \: \: \: \: x _{2}- 4 = - 6 \\ x_{1} = 6 + 4 \: \: \: \: \: \: \: \: \: \: x_{2} = - 6 + 4 \\ x_{1} = 10 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x_{2} = - 2[/tex]
2)
Well he started off correct to the point of completing the square.
[tex](x - 3) {}^{2} = 16 \\ x - 3 = ±4 \\ \: x_{1} - 3 = 4 \: \: \: \: \: \: \: \: \: \: \: \: x_{2} - 3 = - 4 \\ x_{1} = 7 \: \: \: \: \: \: \: \: \: \: x_{2} = - 1[/tex]
