(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

Rephrasing the equation: First, let's rephrase the ＂What are the solutions to the equation (x+3)² - 4 = 0?＂Isolating the squared term: Now, let's solve the equation step by step. The first step is to isolate the squared term by adding 4 to both sides of the equation: (x+3)² - 4 + 4 = 0 + 4 (x+3)² = 4Taking 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: √((x+3)²) = ±√4 x + 3 = ±2Solving the first equation: Now we have two separate equations to solve for x: 1. x + 3 = 2 2. x + 3 = -2Solving the second equation: Let's solve the first equation: x + 3 = 2 Subtract 3 from both sides: x = 2 - 3 x = -1Finding the solutions: Now let's solve the second equation: x + 3 = -2 Subtract 3 from both sides: x = -2 - 3 x = -5We have found two solutions to the equation (x+3)² - 4 = 0: x = -1 and x = -5

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

(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

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

View step-by-step help

Home

Math Problems

Algebra 1

Compare linear and exponential growth

Full solution

(x+3)^{2}-4=0

\newline

What are the solutions to the given equation?

\newline

Choose

1

answer:

\newline

(A)

x=1, x=5

\newline

(B)

x=1, x=-5

\newline

(C)

x=-1, x=5

\newline

(D)

x=-1, x=-5

Rephrasing the equation: First, let's rephrase the ＂What are the solutions to the equation $(x+3)^2 - 4 = 0$ ?＂
Isolating the squared term: Now, let's solve the equation step by step. The first step is to isolate the squared term by adding $4$ to both sides of the equation: $\newline$ $(x+3)^2 - 4 + 4 = 0 + 4$ $\newline$ $(x+3)^2 = 4$
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: $\newline$ $\sqrt{((x+3)^2)} = \pm\sqrt{4}$ $\newline$ $x + 3 = \pm2$
Solving the first equation: Now we have two separate equations to solve for x: $\newline$ $1$ . $x + 3 = 2$ $\newline$ $2$ . $x + 3 = -2$
Solving the second equation: Let's solve the first equation: $\newline$ $x + 3 = 2$ $\newline$ Subtract $3$ from both sides: $\newline$ $x = 2 - 3$ $\newline$ $x = -1$
Finding the solutions: Now let's solve the second equation: $\newline$ $x + 3 = -2$ $\newline$ Subtract $3$ from both sides: $\newline$ $x = -2 - 3$ $\newline$ $x = -5$
Finding the solutions: Now let's solve the second equation: $\newline$ $x + 3 = -2$ $\newline$ Subtract $3$ from both sides: $\newline$ $x = -2 - 3$ $\newline$ $x = -5$ We have found two solutions to the equation $(x+3)^2 - 4 = 0$ : $\newline$ $x = -1$ and $x = -5$