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

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

Find the zeros of 4=x^2+1

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

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4 = x^{2} + 1

x = 3, x = - 3

Decimal Notation

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

Solution steps

4 = x^{2} + 1

Switch sides

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

y^{4} + 8 y^{2} + 16

r^{2} - 36

- 14 a - a^{2} = 0

(x - 2) + (2 x - 1) (x - 3) = x (3 x - 3) - 2 x

\frac{9 c}{10} = \frac{9}{5}

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