What are the solutions to the equation (-12/13)^2+x^2=1 ?

The solutions to the equation (-12/13)^2+x^2=1 are x= 5/13 ,x=-5/13

Find the zeros of (-12/13)^2+x^2=1

The zeros of (-12/13)^2+x^2=1 are x= 5/13 ,x=-5/13

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(- \frac{12}{13})^{2} + x^{2} = 1

x = \frac{5}{13}, x = - \frac{5}{13}

Decimal Notation

x = 0.38461 \dots, x = - 0.38461 \dots

Solution steps

(- \frac{12}{13})^{2} + x^{2} = 1

Expand

(- \frac{12}{13})^{2} + x^{2} : \frac{144}{169} + x^{2}

\frac{144}{169} + x^{2} = 1

Move

\frac{144}{169}

to the right side

x^{2} = 1 - \frac{144}{169}

Simplify

1 - \frac{144}{169} : \frac{25}{169}

For

x^{2} = f (a)

the solutions are

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

x = \frac{25}{169}, x = - \frac{25}{169}

\frac{25}{169} = \frac{5}{13}

- \frac{25}{169} = - \frac{5}{13}

x = \frac{5}{13}, x = - \frac{5}{13}

\frac{1}{2} x^{2} + 7 x = 0

x^{2} - 3 x - 44 = - 3 x + 5 - 3 x + 5

y, x^{4} - 7 x y^{2} + y = 7

x^{2} + (\frac{5}{13})^{2} = 1

x, f = 3 x^{2} + 2 y + 5

What are the solutions to the equation (-12/13)^2+x^2=1 ?
The solutions to the equation (-12/13)^2+x^2=1 are x= 5/13 ,x=-5/13
Find the zeros of (-12/13)^2+x^2=1
The zeros of (-12/13)^2+x^2=1 are x= 5/13 ,x=-5/13