What are the solutions to the equation (x+3)^2=1 ?

The solutions to the equation (x+3)^2=1 are x=-2,x=-4

Find the zeros of (x+3)^2=1

The zeros of (x+3)^2=1 are x=-2,x=-4

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(x + 3)^{2} = 1

x = - 2, x = - 4

Solution steps

(x + 3)^{2} = 1

For

(g (x))^{2} = f (a)

the solutions are

g (x) = f (a), - f (a)

Solve

x + 3 = 1 : x = - 2

Solve

x + 3 = - 1 : x = - 4

The solutions to the quadratic equation are:

x = - 2, x = - 4

3 (60)

\frac{( 3 k ^{4} ) ^{5}}{- 15 k ^{10}}

5 (\frac{1}{2} x + 3) \leq 4 (\frac{1}{3} x + 5)

3 a^{4} b - 192 ab

x + 18 \geq 8 x + 4

What are the solutions to the equation (x+3)^2=1 ?
The solutions to the equation (x+3)^2=1 are x=-2,x=-4
Find the zeros of (x+3)^2=1
The zeros of (x+3)^2=1 are x=-2,x=-4