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

The solutions to the equation solvefor x,2x^2-5xy+2y^2=0 are x=2y,x= y/2

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

The zeros of solvefor x,2x^2-5xy+2y^2=0 are x=2y,x= y/2

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solve for

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

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

Solution steps

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

x^{2} - 2 x - 52 = 3 x - 2

(x - 5) (x - 3) = 15

y, y^{2} = x^{3} + x + 17

x, f (x + 3) = x^{2} - 1

(5 x - 1) (x + 7) = 0

What are the solutions to the equation solvefor x,2x^2-5xy+2y^2=0 ?
The solutions to the equation solvefor x,2x^2-5xy+2y^2=0 are x=2y,x= y/2
Find the zeros of solvefor x,2x^2-5xy+2y^2=0
The zeros of solvefor x,2x^2-5xy+2y^2=0 are x=2y,x= y/2