Questions attached below (❁´◡`❁)

Problem 2

Josh forgot to apply the square root to 16 when he went from to

Also, he forgot about the plus/minus.

This is what his steps should look like

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...

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.

idk

step-by-step explanation: