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

The solutions to the equation x^2+(-2^2)=1 are x=sqrt(5),x=-sqrt(5)

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

The zeros of x^2+(-2^2)=1 are x=sqrt(5),x=-sqrt(5)

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

x = 5, x = - 5

Decimal Notation

x = 2.23606 \dots, x = - 2.23606 \dots

Solution steps

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

Expand

x^{2} + (- 2^{2}) : x^{2} - 4

x^{2} - 4 = 1

Move

4

to the right side

x^{2} = 5

For

x^{2} = f (a)

the solutions are

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

x = 5, x = - 5

x^{2} - 2 = 3 x^{2} + 4 x

(x - 2)^{2} - 3 = - 3

(4 - 3 x)^{2} = x

9 x (x - 4) = 2 x (x - 4)

4 - x^{2} = 5 x

What are the solutions to the equation x^2+(-2^2)=1 ?
The solutions to the equation x^2+(-2^2)=1 are x=sqrt(5),x=-sqrt(5)
Find the zeros of x^2+(-2^2)=1
The zeros of x^2+(-2^2)=1 are x=sqrt(5),x=-sqrt(5)