What are the solutions to the equation solvefor x,x^2+2xy+3y^2=15 ?

The solutions to the equation solvefor x,x^2+2xy+3y^2=15 are x=-y+sqrt(-2y^2+15),x=-y-sqrt(-2y^2+15)

Find the zeros of solvefor x,x^2+2xy+3y^2=15

The zeros of solvefor x,x^2+2xy+3y^2=15 are x=-y+sqrt(-2y^2+15),x=-y-sqrt(-2y^2+15)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solve for

x, x^{2} + 2 x y + 3 y^{2} = 15

x = - y + - 2 y^{2} + 15, x = - y - - 2 y^{2} + 15

Solution steps

x^{2} + 2 x y + 3 y^{2} = 15

Move

15

to the left side

x^{2} + 2 x y + 3 y^{2} - 15 = 0

Write in the standard form

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

x^{2} + 2 y x + 3 y^{2} - 15 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 y \pm ( 2 y ) ^{2} - 4 \cdot 1 \cdot ( 3 y ^{2} - 15 )}{2 \cdot 1}

Simplify

(2 y)^{2} - 4 \cdot 1 \cdot (3 y^{2} - 15) : 2 - 2 y^{2} + 15

x_{1, 2} = \frac{- 2 y \pm 2 - 2 y ^{2} + 15}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 y + 2 - 2 y ^{2} + 15}{2 \cdot 1}, x_{2} = \frac{- 2 y - 2 - 2 y ^{2} + 15}{2 \cdot 1}

x = \frac{- 2 y + 2 - 2 y ^{2} + 15}{2 \cdot 1} : - y + - 2 y^{2} + 15

x = \frac{- 2 y - 2 - 2 y ^{2} + 15}{2 \cdot 1} : - y - - 2 y^{2} + 15

The solutions to the quadratic equation are:

x = - y + - 2 y^{2} + 15, x = - y - - 2 y^{2} + 15

x, x^{2} + y^{2} + 4 y = - 4

2 x^{2} + 16 x - 12 = 0

- 17 x^{2} + 14 + 6 x = 6 x - 7 x^{2}

y, z = (1 - 2 y) (3 y + 2)

4 x^{2} + 5 x + 10 = 0

What are the solutions to the equation solvefor x,x^2+2xy+3y^2=15 ?
The solutions to the equation solvefor x,x^2+2xy+3y^2=15 are x=-y+sqrt(-2y^2+15),x=-y-sqrt(-2y^2+15)
Find the zeros of solvefor x,x^2+2xy+3y^2=15
The zeros of solvefor x,x^2+2xy+3y^2=15 are x=-y+sqrt(-2y^2+15),x=-y-sqrt(-2y^2+15)