What are the solutions to the equation 52^2+48^2=x^2 ?

The solutions to the equation 52^2+48^2=x^2 are x=4sqrt(313),x=-4sqrt(313)

Find the zeros of 52^2+48^2=x^2

The zeros of 52^2+48^2=x^2 are x=4sqrt(313),x=-4sqrt(313)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5 2^{2} + 4 8^{2} = x^{2}

x = 4313, x = - 4313

Decimal Notation

x = 70.76722 \dots, x = - 70.76722 \dots

Solution steps

5 2^{2} + 4 8^{2} = x^{2}

Switch sides

x^{2} = 5 2^{2} + 4 8^{2}

Simplify

5 2^{2} + 4 8^{2} : 5008

For

x^{2} = f (a)

the solutions are

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

x = 5008, x = - 5008

5008 = 4313

- 5008 = - 4313

x = 4313, x = - 4313

j^{2} - 4 j - 45 = 0

2 j^{2} - 3 j = - 1

14 y^{2} - 112 y = 0

- 3 x^{2} + 5 x - 6 = 0

10 x^{2} + 2 x = 9

What are the solutions to the equation 52^2+48^2=x^2 ?
The solutions to the equation 52^2+48^2=x^2 are x=4sqrt(313),x=-4sqrt(313)
Find the zeros of 52^2+48^2=x^2
The zeros of 52^2+48^2=x^2 are x=4sqrt(313),x=-4sqrt(313)