What are the solutions to the equation (1+x)^2=4 ?

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

Find the zeros of (1+x)^2=4

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

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(1 + x)^{2} = 4

x = 1, x = - 3

Solution steps

(1 + x)^{2} = 4

For

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

the solutions are

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

Solve

1 + x = 4 : x = 1

Solve

1 + x = - 4 : x = - 3

The solutions to the quadratic equation are:

x = 1, x = - 3

(x - 5) (x - 6) = 0

5 x^{2} = 10 x + 4

- 18.6 x^{2} + 497 x - 3010 = 0

x^{2} - 4 x + \frac{15}{4} = 0

m^{2} - 12 m + 27 = 0

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