What are the solutions to the equation x^2+(1/6)^2=1^2 ?

The solutions to the equation x^2+(1/6)^2=1^2 are x=(sqrt(35))/6 ,x=-(sqrt(35))/6

Find the zeros of x^2+(1/6)^2=1^2

The zeros of x^2+(1/6)^2=1^2 are x=(sqrt(35))/6 ,x=-(sqrt(35))/6

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x^{2} + (\frac{1}{6})^{2} = 1^{2}

x = \frac{35}{6}, x = - \frac{35}{6}

Decimal Notation

x = 0.98601 \dots, x = - 0.98601 \dots

Solution steps

x^{2} + (\frac{1}{6})^{2} = 1^{2}

Expand

x^{2} + (\frac{1}{6})^{2} : x^{2} + \frac{1}{36}

Expand

1^{2} : 1

x^{2} + \frac{1}{36} = 1

Move

\frac{1}{36}

to the right side

x^{2} = 1 - \frac{1}{36}

Simplify

1 - \frac{1}{36} : \frac{35}{36}

For

x^{2} = f (a)

the solutions are

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

x = \frac{35}{36}, x = - \frac{35}{36}

\frac{35}{36} = \frac{35}{6}

- \frac{35}{36} = - \frac{35}{6}

x = \frac{35}{6}, x = - \frac{35}{6}

4 x^{3} + 35 x^{2} + 88 x + 48

4 (x - 3) - 7 < 2 (x - 3)

(64)^{\frac{5}{6}}

12 x^{2} - 13 x - 4

\frac{2}{2 + 1}

What are the solutions to the equation x^2+(1/6)^2=1^2 ?
The solutions to the equation x^2+(1/6)^2=1^2 are x=(sqrt(35))/6 ,x=-(sqrt(35))/6
Find the zeros of x^2+(1/6)^2=1^2
The zeros of x^2+(1/6)^2=1^2 are x=(sqrt(35))/6 ,x=-(sqrt(35))/6