What are the solutions to the equation x^2+(-2/7)^2=1 ?

The solutions to the equation x^2+(-2/7)^2=1 are x=(3sqrt(5))/7 ,x=-(3sqrt(5))/7

Find the zeros of x^2+(-2/7)^2=1

The zeros of x^2+(-2/7)^2=1 are x=(3sqrt(5))/7 ,x=-(3sqrt(5))/7

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x^{2} + (- \frac{2}{7})^{2} = 1

x = \frac{3 5}{7}, x = - \frac{3 5}{7}

Decimal Notation

x = 0.95831 \dots, x = - 0.95831 \dots

Solution steps

x^{2} + (- \frac{2}{7})^{2} = 1

Expand

x^{2} + (- \frac{2}{7})^{2} : x^{2} + \frac{4}{49}

x^{2} + \frac{4}{49} = 1

Move

\frac{4}{49}

to the right side

x^{2} = 1 - \frac{4}{49}

Simplify

1 - \frac{4}{49} : \frac{45}{49}

For

x^{2} = f (a)

the solutions are

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

x = \frac{45}{49}, x = - \frac{45}{49}

\frac{45}{49} = \frac{3 5}{7}

- \frac{45}{49} = - \frac{3 5}{7}

x = \frac{3 5}{7}, x = - \frac{3 5}{7}

\frac{5 x + 5}{6} - 1 < \frac{19 x + 1}{3} - \frac{8 x + 1}{4}

- 2 x^{2} + 4 x

w^{22}

\frac{2 x}{3} = 3 + \frac{x}{2}

- 3 x^{- 4} y^{0}

What are the solutions to the equation x^2+(-2/7)^2=1 ?
The solutions to the equation x^2+(-2/7)^2=1 are x=(3sqrt(5))/7 ,x=-(3sqrt(5))/7
Find the zeros of x^2+(-2/7)^2=1
The zeros of x^2+(-2/7)^2=1 are x=(3sqrt(5))/7 ,x=-(3sqrt(5))/7