What are the solutions to the equation x^2+(-3/7)^2=1 ?

The solutions to the equation x^2+(-3/7)^2=1 are x=(2sqrt(10))/7 ,x=-(2sqrt(10))/7

Find the zeros of x^2+(-3/7)^2=1

The zeros of x^2+(-3/7)^2=1 are x=(2sqrt(10))/7 ,x=-(2sqrt(10))/7

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x^{2} + (- \frac{3}{7})^{2} = 1

x = \frac{2 10}{7}, x = - \frac{2 10}{7}

Decimal Notation

x = 0.90350 \dots, x = - 0.90350 \dots

Solution steps

x^{2} + (- \frac{3}{7})^{2} = 1

Expand

x^{2} + (- \frac{3}{7})^{2} : x^{2} + \frac{9}{49}

x^{2} + \frac{9}{49} = 1

Move

\frac{9}{49}

to the right side

x^{2} = 1 - \frac{9}{49}

Simplify

1 - \frac{9}{49} : \frac{40}{49}

For

x^{2} = f (a)

the solutions are

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

x = \frac{40}{49}, x = - \frac{40}{49}

\frac{40}{49} = \frac{2 10}{7}

- \frac{40}{49} = - \frac{2 10}{7}

x = \frac{2 10}{7}, x = - \frac{2 10}{7}

\frac{8}{11} x^{2} - \frac{2}{11} x = 0

5 x^{2} = - 20

\frac{x ( x - 5 )}{6} - 1 = 0

9 x^{2} + 6 x - 11 = 0

x, 0 = x^{2} + y^{2}

What are the solutions to the equation x^2+(-3/7)^2=1 ?
The solutions to the equation x^2+(-3/7)^2=1 are x=(2sqrt(10))/7 ,x=-(2sqrt(10))/7
Find the zeros of x^2+(-3/7)^2=1
The zeros of x^2+(-3/7)^2=1 are x=(2sqrt(10))/7 ,x=-(2sqrt(10))/7