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(x+3)^(2)-4=0 What are the solutions to the given equation? Choose 1 answer: (A) x=1,x=5 (B) x=1,x=-5 (c) x=-1,x=5 (D) x=-1,x=-5

(x+3)2−4=0 (x+3)^{2}-4=0 \newlineWhat are the solutions to the given equation?\newlineChoose 11 answer:\newline(A) x=1,x=5 x=1, x=5 \newline(B) x=1,x=−5 x=1, x=-5 \newline(C) x=−1,x=5 x=-1, x=5 \newline(D) x=−1,x=−5 x=-1, x=-5

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Q. (x+3)2−4=0 (x+3)^{2}-4=0 \newlineWhat are the solutions to the given equation?\newlineChoose 11 answer:\newline(A) x=1,x=5 x=1, x=5 \newline(B) x=1,x=−5 x=1, x=-5 \newline(C) x=−1,x=5 x=-1, x=5 \newline(D) x=−1,x=−5 x=-1, x=-5
  1. Rephrasing the equation: First, let's rephrase the ＂What are the solutions to the equation (x+3)2−4=0(x+3)^2 - 4 = 0?＂
  2. Isolating the squared term: Now, let's solve the equation step by step. The first step is to isolate the squared term by adding 44 to both sides of the equation:\newline(x+3)2−4+4=0+4(x+3)^2 - 4 + 4 = 0 + 4\newline(x+3)2=4(x+3)^2 = 4
  3. Taking the square root: Next, we take the square root of both sides of the equation to solve for x. Remember that taking the square root of a squared expression gives us two possible solutions, one positive and one negative:\newline((x+3)2)=±4\sqrt{((x+3)^2)} = \pm\sqrt{4}\newlinex+3=±2x + 3 = \pm2
  4. Solving the first equation: Now we have two separate equations to solve for x:\newline11. x+3=2x + 3 = 2\newline22. x+3=−2x + 3 = -2
  5. Solving the second equation: Let's solve the first equation:\newlinex+3=2x + 3 = 2\newlineSubtract 33 from both sides:\newlinex=2−3x = 2 - 3\newlinex=−1x = -1
  6. Finding the solutions: Now let's solve the second equation:\newlinex+3=−2x + 3 = -2\newlineSubtract 33 from both sides:\newlinex=−2−3x = -2 - 3\newlinex=−5x = -5
  7. Finding the solutions: Now let's solve the second equation:\newlinex+3=−2x + 3 = -2\newlineSubtract 33 from both sides:\newlinex=−2−3x = -2 - 3\newlinex=−5x = -5We have found two solutions to the equation (x+3)2−4=0(x+3)^2 - 4 = 0:\newlinex=−1x = -1 and x=−5x = -5

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