(x+1)^(2)-36=0 What are the solutions to the given equation? Choose 1 answer: (A) x=-5,x=7 (B) x=-5,x=-7 (c) x=5,x=-7 (D) x=5,x=7

Set Equation to Zero: Set the equation equal to zero. (x+1)² - 36 = 0 This is a quadratic equation in the form of a difference of squares, which can be factored.Factor the Equation: Factor the equation. (x+1)² - 36 = (x+1 + 6)(x+1 - 6) = 0 This is because a² - b² = (a + b)(a - b), where a = (x+1) and b = 6.Solve for x: Set each factor equal to zero and solve for x. (x+1 + 6) = 0 or (x+1 - 6) = 0 x + 7 = 0 or x - 5 = 0Final Solutions: Solve each equation for x. x + 7 = 0 gives x = -7 x - 5 = 0 gives x = 5

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(x+1)^(2)-36=0
What are the solutions to the given equation?
Choose 1 answer:
(A)
x=-5,x=7
(B)
x=-5,x=-7
(c)
x=5,x=-7
(D)
x=5,x=7

$(x+1)^{2}-36=0$ $\newline$ What are the solutions to the given equation? $\newline$ Choose $1$ answer: $\newline$ (A) $x=-5, x=7$ $\newline$ (B) $x=-5, x=-7$ $\newline$ (C) $x=5, x=-7$ $\newline$ (D) $x=5, x=7$

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Algebra 1

Solve linear equations: mixed review

Full solution

(x+1)^{2}-36=0

\newline

What are the solutions to the given equation?

\newline

Choose

1

answer:

\newline

(A)

x=-5, x=7

\newline

(B)

x=-5, x=-7

\newline

(C)

x=5, x=-7

\newline

(D)

x=5, x=7

Set Equation to Zero: Set the equation equal to zero. $\newline$ $(x+1)^2 - 36 = 0$ $\newline$ This is a quadratic equation in the form of a difference of squares, which can be factored.
Factor the Equation: Factor the equation. $\newline$ $(x+1)^2 - 36 = (x+1 + 6)(x+1 - 6) = 0$ $\newline$ This is because $a^2 - b^2 = (a + b)(a - b)$ , where $a = (x+1)$ and $b = 6$ .
Solve for x: Set each factor equal to zero and solve for x. $\newline$ $(x+1 + 6) = 0$ or $(x+1 - 6) = 0$ $\newline$ $x + 7 = 0$ or $x - 5 = 0$
Final Solutions: Solve each equation for $x$ . $\newline$ $x + 7 = 0$ gives $x = -7$ $\newline$ $x - 5 = 0$ gives $x = 5$