What are the solutions to the equation r^2=(sqrt(3))^2+(-1)^2 ?

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

Find the zeros of r^2=(sqrt(3))^2+(-1)^2

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

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r^{2} = (3)^{2} + (- 1)^{2}

r = 2, r = - 2

Solution steps

r^{2} = (3)^{2} + (- 1)^{2}

Simplify

(3)^{2} + (- 1)^{2} : 4

For

x^{2} = f (a)

the solutions are

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

r = 4, r = - 4

4 = 2

- 4 = - 2

r = 2, r = - 2

(4 x + 3)^{2} = 18

0.02 (0.5 x^{2} - x) = 0, 25

x, zy = x^{2}

x^{2} - 7 x = - 3

5 p^{2} - 14 p = 24

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