Question

Hi! can anyone help me pls?

Answer 1

-4 and 4

Explanation:

All you need to do is solve for y

`y^(2) - 1 = 15\\` (plus 1 to both sides)

`y^(2) = 16` (take square root of 16)

`y = √(16)`

`y = +-4` (whenever you to the root of a perfect square you want the positive and negative root)

To double check if your answer is right: plug in -4 and +4 in as y and see if it equals 15

1.
`(4)^(2) - 1 =15`

`16-1 = 15\\15= 15`

(positive 4 checks out)

2.
`(-4)^(2) - 1 = 15`

`16 - 1 = 15\\15 = 15`

( negative 4 checks out)

Answer 2

Option B, -4 and 4

Explanation:

Step 1: Subtract 15 from both sides

y^2 - 1 - 15 = 15 - 15

y^2 - 16 = 0

Step 2: Factor

y^2 - 16

(y + 4)(y - 4)

Step 3: Find the solutions

y + 4 - 4 = 0 - 4 and y - 4 + 4 = 0 + 4

y = -4 and y = 4

Answer: Option B, -4 and 4