What is the answer to solvefor t,x=sqrt(t^2+1) ?

The answer to solvefor t,x=sqrt(t^2+1) is t=sqrt(x^2-1) {x>= 1},t=-sqrt(x^2-1) {x>= 1}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor t,x=sqrt(t^2+1)

solve for

t, x = t^{2} + 1

t = x^{2} - 1 {x \geq 1}, t = - x^{2} - 1 {x \geq 1}

Solution steps

x = t^{2} + 1

Square both sides:

x^{2} = t^{2} + 1

x^{2} = t^{2} + 1

Solve

x^{2} = t^{2} + 1 : t = x^{2} - 1, t = - x^{2} - 1

t = x^{2} - 1, t = - x^{2} - 1

Verify Solutions:

t = x^{2} - 1 {x \geq 1}, t = - x^{2} - 1 {x \geq 1}

The solutions are

t = x^{2} - 1 {x \geq 1}, t = - x^{2} - 1 {x \geq 1}

- 4 x - 1 + - 2 x - 1 = 1

x + x + 1 = 5

29.5 = a^{\frac{3}{2}}

x^{1.25} = 20

What is the answer to solvefor t,x=sqrt(t^2+1) ?
The answer to solvefor t,x=sqrt(t^2+1) is t=sqrt(x^2-1) {x>= 1},t=-sqrt(x^2-1) {x>= 1}