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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 (t-10)^{2}-16=0 \newlineWhat are the solutions to the given equation?\newlineChoose 11 answer:\newline(A) t=4 t=4 and t=36 t=36 \newline(B) t=6 t=6 and t=14 t=14 \newline(C) t=10 t=10 and t=−16 t=-16 \newline(D) t=10 t=10 and t=16 t=16

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Q. (t−10)2−16=0 (t-10)^{2}-16=0 \newlineWhat are the solutions to the given equation?\newlineChoose 11 answer:\newline(A) t=4 t=4 and t=36 t=36 \newline(B) t=6 t=6 and t=14 t=14 \newline(C) t=10 t=10 and t=−16 t=-16 \newline(D) t=10 t=10 and t=16 t=16
  1. Recognize the equation form: Recognize that the equation is in the form of a difference of squares, which can be factored as (a2−b2)=(a+b)(a−b)(a^2 - b^2) = (a + b)(a - b).\newlineThe equation (t−10)2−16=0(t-10)^2 - 16 = 0 can be rewritten as (t−10)2−42=0(t-10)^2 - 4^2 = 0.
  2. Factor using difference of squares: Factor the equation using the difference of squares formula.\newline(t−10+4)(t−10−4)=0(t-10 + 4)(t-10 - 4) = 0\newline(t−6)(t−14)=0(t-6)(t-14) = 0
  3. Solve for t: Set each factor equal to zero and solve for t.\newlinet−6=0t - 6 = 0 or t−14=0t - 14 = 0\newlinet=6t = 6 or t=14t = 14
  4. Check solutions: Check the solutions to ensure they satisfy the original equation.\newlineFor t=6t = 6: (6−10)2−16=(−4)2−16=16−16=0(6-10)^2 - 16 = (-4)^2 - 16 = 16 - 16 = 0\newlineFor t=14t = 14: (14−10)2−16=(4)2−16=16−16=0(14-10)^2 - 16 = (4)^2 - 16 = 16 - 16 = 0\newlineBoth solutions satisfy the original equation.

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