Question

Questions attached below (❁´◡`❁)

Answer 1

Problem 2

Josh forgot to apply the square root to 16 when he went from
`(x-3)^2 = 16` to
`x-3 = 16`

Also, he forgot about the plus/minus.

This is what his steps should look like

`x^2 - 6x - 7 = 0\\\\x^2 - 6x = 7\\\\x^2 - 6x +9= 7+9\\\\(x-3)^2= 16\\\\x-3= \pm√(16)\\\\x-3= 4 \text{ or } x-3= -4\\\\x= 7 \text{ or } x= -1\\\\`

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

====================================================

Problem 3

We will have

a = 1, b = -4, c = 3

plugged into the quadratic formula below to get...

`x = (-b\pm√(b^2-4ac))/(2a)\\\\x = (-(-4)\pm√((-4)^2-4(1)(3)))/(2(1))\\\\x = (4\pm√(4))/(2)\\\\x = (4\pm2)/(2)\\\\x = (4+2)/(2) \ \text{ or } \ x = (4-2)/(2)\\\\x = (6)/(2) \ \text{ or } \ x = (2)/(2)\\\\x = 3 \ \text{ or } \ x = 1\\\\`

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.