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

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

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

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

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(x + 2) (x + 2) = 1

x = - 1, x = - 3

Solution steps

(x + 2) (x + 2) = 1

Simplify

(x + 2)^{2} = 1

For

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

the solutions are

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

Solve

x + 2 = 1 : x = - 1

Solve

x + 2 = - 1 : x = - 3

The solutions to the quadratic equation are:

x = - 1, x = - 3

\frac{x}{6} = 7

- x^{2} + 2 x > 0

4 x \geq 5 x - 7

64 - 81 d^{2}

∣ 5 x - 2 ∣ \leq 8

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