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

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

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

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

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x^{2} + (1)^{2} = 4

x = 3, x = - 3

Decimal Notation

x = 1.73205 \dots, x = - 1.73205 \dots

Solution steps

x^{2} + (1)^{2} = 4

Expand

x^{2} + 1^{2} : x^{2} + 1

x^{2} + 1 = 4

Move

1

to the right side

x^{2} = 3

For

x^{2} = f (a)

the solutions are

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

x = 3, x = - 3

5 z^{2} - 35 = 0

\frac{1}{2} x^{2} - 18 = 0

a, f = \frac{( x - a ) ^{2}}{( x - b ) ^{2}}

12 x^{2} - 42 x + 28 = 0

3 x^{2} + 12 x - 12 = 0

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