(t-10)^(2)-16=0 What are the solutions to the given equation? Choose 1 answer: (A) t=4 and t=36 (B) t=6 and t=14 (C) t=10 and t=-16 (D) t=10 and t=16

Recognize the equation form: Recognize that the equation is in the form of a difference of squares, which can be factored as (a^2 - b^2) = (a + b)(a - b). The equation (t-10)² - 16 = 0 can be rewritten as (t-10)² - 4² = 0.Factor using difference of squares: Factor the equation using the difference of squares formula. (t-10 + 4)(t-10 - 4) = 0 (t-6)(t-14) = 0Solve for t: Set each factor equal to zero and solve for t. t - 6 = 0 or t - 14 = 0 t = 6 or t = 14Check solutions: Check the solutions to ensure they satisfy the original equation. For t = 6: (6-10)² - 16 = (-4)² - 16 = 16 - 16 = 0 For t = 14: (14-10)² - 16 = (4)² - 16 = 16 - 16 = 0 Both solutions satisfy the original equation.

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(t-10)^(2)-16=0
What are the solutions to the given equation?
Choose 1 answer:
(A)
t=4 and
t=36
(B)
t=6 and
t=14
(C)
t=10 and
t=-16
(D)
t=10 and
t=16

$(t-10)^{2}-16=0$ $\newline$ What are the solutions to the given equation? $\newline$ Choose $1$ answer: $\newline$ (A) $t=4$ and $t=36$ $\newline$ (B) $t=6$ and $t=14$ $\newline$ (C) $t=10$ and $t=-16$ $\newline$ (D) $t=10$ and $t=16$

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Algebra 1

Compare linear and exponential growth

Full solution

(t-10)^{2}-16=0

\newline

What are the solutions to the given equation?

\newline

Choose

1

answer:

\newline

(A)

t=4

and

t=36

\newline

(B)

t=6

and

t=14

\newline

(C)

t=10

and

t=-16

\newline

(D)

t=10

and

t=16

Recognize the equation form: Recognize that the equation is in the form of a difference of squares, which can be factored as $(a^2 - b^2) = (a + b)(a - b)$ . $\newline$ The equation $(t-10)^2 - 16 = 0$ can be rewritten as $(t-10)^2 - 4^2 = 0$ .
Factor using difference of squares: Factor the equation using the difference of squares formula. $\newline$ $(t-10 + 4)(t-10 - 4) = 0$ $\newline$ $(t-6)(t-14) = 0$
Solve for t: Set each factor equal to zero and solve for t. $\newline$ $t - 6 = 0$ or $t - 14 = 0$ $\newline$ $t = 6$ or $t = 14$
Check solutions: Check the solutions to ensure they satisfy the original equation. $\newline$ For $t = 6$ : $(6-10)^2 - 16 = (-4)^2 - 16 = 16 - 16 = 0$ $\newline$ For $t = 14$ : $(14-10)^2 - 16 = (4)^2 - 16 = 16 - 16 = 0$ $\newline$ Both solutions satisfy the original equation.