(x+3)^(2)-49=0 What are the solutions to the given equation? Choose 1 answer: (A) x=-4,x=-10 (B) x=4,x=-10 (c) x=4,x=10 (D) x=-4,x=10

Simplify the factors: Next, we simplify the factors: (x+10)(x-4) = 0 This gives us two separate equations to solve for x: x+10 = 0 or x-4 = 0Solve the first equation: Now we solve each equation for x: For the first equation: x+10 = 0 x = -10 For the second equation: x-4 = 0 x = 4Solve the second equation: We have found the two solutions to the equation: x = -10 and x = 4 These correspond to answer choice (B) x=4, x=-10.

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(x+3)^(2)-49=0
What are the solutions to the given equation?
Choose 1 answer:
(A)
x=-4,x=-10
(B)
x=4,x=-10
(c)
x=4,x=10
(D)
x=-4,x=10

$(x+3)^{2}-49=0$ $\newline$ What are the solutions to the given equation? $\newline$ Choose $1$ answer: $\newline$ (A) $x=-4, x=-10$ $\newline$ (B) $x=4, x=-10$ $\newline$ (C) $x=4, x=10$ $\newline$ (D) $x=-4, x=10$

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Algebra 1

Solve linear equations: mixed review

Full solution

(x+3)^{2}-49=0

\newline

What are the solutions to the given equation?

\newline

Choose

1

answer:

\newline

(A)

x=-4, x=-10

\newline

(B)

x=4, x=-10

\newline

(C)

x=4, x=10

\newline

(D)

x=-4, x=10

Simplify the factors: Next, we simplify the factors: $\newline$ $(x+10)(x-4) = 0$ $\newline$ This gives us two separate equations to solve for $x$ : $\newline$ $x+10 = 0$ or $x-4 = 0$
Solve the first equation: Now we solve each equation for $x$ :
For the first equation:
$x+10 = 0$
$x = -10$
For the second equation:
$x-4 = 0$
$x = 4$
Solve the second equation: We have found the two solutions to the equation: $\newline$ $x = -10$ and $x = 4$ $\newline$ These correspond to answer choice (B) $x=4$ , $x=-10$ .