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

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

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

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

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3^{2} = x^{2} + 1^{2}

x = 22, x = - 22

Decimal Notation

x = 2.82842 \dots, x = - 2.82842 \dots

Solution steps

3^{2} = x^{2} + 1^{2}

Switch sides

x^{2} + 1^{2} = 3^{2}

Move

1^{2}

to the right side

x^{2} = 8

For

x^{2} = f (a)

the solutions are

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

x = 8, x = - 8

8 = 22

- 8 = - 22

x = 22, x = - 22

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

3 x^{2} - 12 x + 5

2 x^{2} - 3 x - 20 = x^{2} + 34

2 x^{2} + 2 = 10

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

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