Create a list of steps, in order, that will solve the following equation. (x+3)^(2)-1=35 Solution steps: Add 1 to both sides Add 3 to both sides Subtract 1 from both sides Subtract 3 from both sides Take the square root of both sides

Add 1 to isolate squared term: Add 1 to both sides to isolate the squared term. We have (x+3)^(2) - 1 = 35. By adding 1 to both sides, we get (x+3)^(2) = 36.Take square root to solve: Take the square root of both sides to solve for x+3. Taking the square root of both sides, we get x+3 = ±√36. The square root of 36 is 6, so we have x+3 = ±6.Subtract 3 to find x: Subtract 3 from both sides to solve for x. We have x+3 = ±6. By subtracting 3 from both sides, we get x = ±6 - 3. This gives us two possible solutions: x = 6 - 3 or x = -6 - 3.

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Create a list of steps, in order, that will solve the following equation.

(x+3)^(2)-1=35
Solution steps:
Add 1 to both sides
Add 3 to both sides
Subtract 1 from both sides
Subtract 3 from both sides
Take the square root of both sides

Create a list of steps, in order, that will solve the following equation. $\newline$ $(x+3)^{2}-1=35$ $\newline$ Solution steps: $\newline$ - Add $1$ to both sides $\newline$ - Add $3$ to both sides $\newline$ - Subtract $1$ from both sides $\newline$ - Subtract $3$ from both sides $\newline$ - Take the square root of both sides

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Math Problems

Algebra 2

Solve a quadratic equation using square roots

Full solution

Q. Create a list of steps, in order, that will solve the following equation.

\newline

(x+3)^{2}-1=35

\newline

Solution steps:

\newline

- Add

1

to both sides

\newline

- Add

3

to both sides

\newline

- Subtract

1

from both sides

\newline

- Subtract

3

from both sides

\newline

- Take the square root of both sides

Add $1$ to isolate squared term: Add $1$ to both sides to isolate the squared term. $\newline$ We have $(x+3)^{2} - 1 = 35$ . By adding $1$ to both sides, we get $(x+3)^{2} = 36$ .
Take square root to solve: Take the square root of both sides to solve for $x+3$ . Taking the square root of both sides, we get $x+3 = \pm\sqrt{36}$ . The square root of $36$ is $6$ , so we have $x+3 = \pm6$ .
Subtract $3$ to find x: Subtract $3$ from both sides to solve for x. $\newline$ We have $x+3 = \pm 6$ . By subtracting $3$ from both sides, we get $x = \pm 6 - 3$ . This gives us two possible solutions: $x = 6 - 3$ or $x = -6 - 3$ .