What are the solutions to the equation solvefor x,2x^2+2y^2z^2-5yzx=0 ?

The solutions to the equation solvefor x,2x^2+2y^2z^2-5yzx=0 are x=2yz,x=(yz)/2

Find the zeros of solvefor x,2x^2+2y^2z^2-5yzx=0

The zeros of solvefor x,2x^2+2y^2z^2-5yzx=0 are x=2yz,x=(yz)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solve for

x, 2 x^{2} + 2 y^{2} z^{2} - 5 yz x = 0

x = 2 yz, x = \frac{yz}{2}

Solution steps

2 x^{2} + 2 y^{2} z^{2} - 5 yz x = 0

Write in the standard form

a x^{2} + b x + c = 0

2 x^{2} - 5 yz x + 2 y^{2} z^{2} = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 5 yz ) \pm ( - 5 yz ) ^{2} - 4 \cdot 2 \cdot 2 y ^{2} z ^{2}}{2 \cdot 2}

Simplify

(- 5 yz)^{2} - 4 \cdot 2 \cdot 2 y^{2} z^{2} : 3 yz

x_{1, 2} = \frac{- ( - 5 yz ) \pm 3 yz}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 5 yz ) + 3 yz}{2 \cdot 2}, x_{2} = \frac{- ( - 5 yz ) - 3 yz}{2 \cdot 2}

x = \frac{- ( - 5 yz ) + 3 yz}{2 \cdot 2} : 2 yz

x = \frac{- ( - 5 yz ) - 3 yz}{2 \cdot 2} : \frac{yz}{2}

The solutions to the quadratic equation are:

x = 2 yz, x = \frac{yz}{2}

2 x^{2} - 12 x + 28

x^{2} + 4 x = 7 x + 4

k^{2} - 3 k - 10 = 0

a^{2} + 5 a + 25 = 0

v, m g h = \frac{1}{2} m v^{2}

What are the solutions to the equation solvefor x,2x^2+2y^2z^2-5yzx=0 ?
The solutions to the equation solvefor x,2x^2+2y^2z^2-5yzx=0 are x=2yz,x=(yz)/2
Find the zeros of solvefor x,2x^2+2y^2z^2-5yzx=0
The zeros of solvefor x,2x^2+2y^2z^2-5yzx=0 are x=2yz,x=(yz)/2