What are the solutions to the equation solvefor x,xy^2-4x^2-3y^2+12x=0 ?

The solutions to the equation solvefor x,xy^2-4x^2-3y^2+12x=0 are x=3,x=(y^2)/4

Find the zeros of solvefor x,xy^2-4x^2-3y^2+12x=0

The zeros of solvefor x,xy^2-4x^2-3y^2+12x=0 are x=3,x=(y^2)/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solve for

x, x y^{2} - 4 x^{2} - 3 y^{2} + 12 x = 0

x = 3, x = \frac{y ^{2}}{4}

Solution steps

x y^{2} - 4 x^{2} - 3 y^{2} + 12 x = 0

Write in the standard form

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

- 4 x^{2} + (y^{2} + 12) x - 3 y^{2} = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( y ^{2} + 12 ) \pm ( y ^{2} + 12 ) ^{2} - 4 ( - 4 ) ( - 3 y ^{2} )}{2 ( - 4 )}

Simplify

(y^{2} + 12)^{2} - 4 (- 4) (- 3 y^{2}) : (y + 23) (y - 23)

x_{1, 2} = \frac{- ( y ^{2} + 12 ) \pm ( y + 2 3 ) ( y - 2 3 )}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- ( y ^{2} + 12 ) + ( y + 2 3 ) ( y - 2 3 )}{2 ( - 4 )}, x_{2} = \frac{- ( y ^{2} + 12 ) - ( y + 2 3 ) ( y - 2 3 )}{2 ( - 4 )}

x = \frac{- ( y ^{2} + 12 ) + ( y + 2 3 ) ( y - 2 3 )}{2 ( - 4 )} : 3

x = \frac{- ( y ^{2} + 12 ) - ( y + 2 3 ) ( y - 2 3 )}{2 ( - 4 )} : \frac{y ^{2}}{4}

The solutions to the quadratic equation are:

x = 3, x = \frac{y ^{2}}{4}

t^{2} - 5 t + 6 = 0

4 x^{2} - 12 x + 4 = 0

x^{2} - 6 x - 5

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

\frac{1}{2} \cdot 2 \cdot 5^{2} + 2 \cdot 9.8 \cdot 20 = \frac{1}{2} \cdot x^{2}

What are the solutions to the equation solvefor x,xy^2-4x^2-3y^2+12x=0 ?
The solutions to the equation solvefor x,xy^2-4x^2-3y^2+12x=0 are x=3,x=(y^2)/4
Find the zeros of solvefor x,xy^2-4x^2-3y^2+12x=0
The zeros of solvefor x,xy^2-4x^2-3y^2+12x=0 are x=3,x=(y^2)/4