Problem 2
Josh forgot to apply the square root to 16 when he went from [tex](x-3)^2 = 16[/tex] to [tex]x-3 = 16[/tex]
Also, he forgot about the plus/minus.
This is what his steps should look like
[tex]x^2 - 6x - 7 = 0\\\\x^2 - 6x = 7\\\\x^2 - 6x +9= 7+9\\\\(x-3)^2= 16\\\\x-3= \pm\sqrt{16}\\\\x-3= 4 \text{ or } x-3= -4\\\\x= 7 \text{ or } x= -1\\\\[/tex]
There are two solutions and they are x = 7 or x = -1. To check each solution, you plug it back into the original equation
Let's try out x = 7
x^2 - 6x - 7 = 0
7^2 - 6(7) - 7 = 0
49 - 42 - 7 = 0
0 = 0 ... solution x = 7 is confirmed. I'll let you check x = -1
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Problem 3
We will have
a = 1, b = -4, c = 3
plugged into the quadratic formula below to get...
[tex]x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(-4)\pm\sqrt{(-4)^2-4(1)(3)}}{2(1)}\\\\x = \frac{4\pm\sqrt{4}}{2}\\\\x = \frac{4\pm2}{2}\\\\x = \frac{4+2}{2} \ \text{ or } \ x = \frac{4-2}{2}\\\\x = \frac{6}{2} \ \text{ or } \ x = \frac{2}{2}\\\\x = 3 \ \text{ or } \ x = 1\\\\[/tex]
The two solutions are x = 3 or x = 1. You would check this by plugging x = 3 back into the original expression x^2 - 4x + 3. The result should be zero. The same applies to x = 1 as well.
