What are the solutions to the equation solvefor x,z=x^2y^2 ?

The solutions to the equation solvefor x,z=x^2y^2 are x=(sqrt(z))/y ,x=-(sqrt(z))/y ;\quad y\ne 0

Find the zeros of solvefor x,z=x^2y^2

The zeros of solvefor x,z=x^2y^2 are x=(sqrt(z))/y ,x=-(sqrt(z))/y ;\quad y\ne 0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solve for

x, z = x^{2} y^{2}

x = \frac{z}{y}, x = - \frac{z}{y}; y \neq = 0

Solution steps

z = x^{2} y^{2}

Switch sides

x^{2} y^{2} = z

Divide both sides by

y^{2}; y \neq = 0

x^{2} = \frac{z}{y ^{2}}; y \neq = 0

For

x^{2} = f (a)

the solutions are

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

x = \frac{z}{y ^{2}}, x = - \frac{z}{y ^{2}}; y \neq = 0

Simplify

\frac{z}{y ^{2}} : \frac{z}{y}

Simplify

- \frac{z}{y ^{2}} : - \frac{z}{y}

x = \frac{z}{y}, x = - \frac{z}{y}; y \neq = 0

(x + 1)^{2} + x^{2} = 25

\frac{x ^{2}}{6} - \frac{5 x}{12} - \frac{1}{4} = 0

3000 (1 + r)^{2} + 2000 (1 + r) = 5895.25

x^{2} + 10 x - 119 = 0

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

What are the solutions to the equation solvefor x,z=x^2y^2 ?
The solutions to the equation solvefor x,z=x^2y^2 are x=(sqrt(z))/y ,x=-(sqrt(z))/y ;\quad y\ne 0
Find the zeros of solvefor x,z=x^2y^2
The zeros of solvefor x,z=x^2y^2 are x=(sqrt(z))/y ,x=-(sqrt(z))/y ;\quad y\ne 0