Question

What is the solution to the square root function P(t) = 8 – 2√t² + 4?

A) t = 1
B) t = 2
C) t = 3
D) t = 4

Answer 1

The solutions to the square root function P(t) = 8 – 2√t² + 4 are:

• B. t = 2
• C. t = 3
• D. t = 4

Step-by-step explanation:

To solve for the solution to the square root function P(t) = 8 – 2√t² + 4, we need to simplify the equation and isolate the square root term.

P(t) = 8 – 2√t² + 4

P(t) = 12 – 2√t²

First, we can combine the constant terms:

P(t) = -2√t² + 12

Next, we can move the constant term to the right side of the equation:

-2√t² = 12 - P(t)

Finally, we can divide both sides by -2 to isolate the square root term:

√t² = (12 - P(t)) / -2

Squaring both sides to eliminate the square root gives:

t² = (12 - P(t))² / 4

Taking the square root of both sides again gives:

t = ±(12 - P(t)) / 2

Since t cannot be negative, we discard the negative solution. Therefore, the solution to the square root function P(t) = 8 – 2√t² + 4 is t = (12 - P(t)) / 2.

Now, let's evaluate the given values of t:

For t = 1:

P(t) = 8 – 2√1² + 4

P(t) = 10

Since P(1) is not equal to the given value of P(t), t = 1 is not a solution.

For t = 2:

P(t) = 8 – 2√2² + 4

P(t) = 8

Since P(2) is equal to the given value of P(t), t = 2 is a solution.

For t = 3:

P(t) = 8 – 2√3² + 4

P(t) = 6

Since P(3) is equal to the given value of P(t), t = 3 is also a solution.

For t = 4:

P(t) = 8 – 2√4² + 4

P(t) = 4

Since P(4) is equal to the given value of P(t), t = 4 is a solution.

Therefore, the solutions to the square root function P(t) = 8 – 2√t² + 4 are:

• B. t = 2
• C. t = 3
• D. t = 4