What are the solutions to the equation (3/4)^2+x^2=1 ?

The solutions to the equation (3/4)^2+x^2=1 are x=(sqrt(7))/4 ,x=-(sqrt(7))/4

Find the zeros of (3/4)^2+x^2=1

The zeros of (3/4)^2+x^2=1 are x=(sqrt(7))/4 ,x=-(sqrt(7))/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\frac{3}{4})^{2} + x^{2} = 1

x = \frac{7}{4}, x = - \frac{7}{4}

Decimal Notation

x = 0.66143 \dots, x = - 0.66143 \dots

Solution steps

(\frac{3}{4})^{2} + x^{2} = 1

Expand

(\frac{3}{4})^{2} + x^{2} : \frac{9}{16} + x^{2}

\frac{9}{16} + x^{2} = 1

Move

\frac{9}{16}

to the right side

x^{2} = 1 - \frac{9}{16}

Simplify

1 - \frac{9}{16} : \frac{7}{16}

For

x^{2} = f (a)

the solutions are

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

x = \frac{7}{16}, x = - \frac{7}{16}

\frac{7}{16} = \frac{7}{4}

- \frac{7}{16} = - \frac{7}{4}

x = \frac{7}{4}, x = - \frac{7}{4}

x^{2} - 5 x + 3 = - 7 x + 11

- 3 x^{2} - 2 x + 5 = 0

(x - 4)^{2} + (x - 3)^{2} = (x - 2)^{2}

- \frac{1}{3} x^{2} + 2 x - 3 = 0

x, y = 2 x^{2}

What are the solutions to the equation (3/4)^2+x^2=1 ?
The solutions to the equation (3/4)^2+x^2=1 are x=(sqrt(7))/4 ,x=-(sqrt(7))/4
Find the zeros of (3/4)^2+x^2=1
The zeros of (3/4)^2+x^2=1 are x=(sqrt(7))/4 ,x=-(sqrt(7))/4