What are the solutions to the equation solvefor x,5x^2-4xy+y^2=0 ?

The solutions to the equation solvefor x,5x^2-4xy+y^2=0 are x=(2y)/5+i y/5 ,x=(2y)/5-i y/5

Find the zeros of solvefor x,5x^2-4xy+y^2=0

The zeros of solvefor x,5x^2-4xy+y^2=0 are x=(2y)/5+i y/5 ,x=(2y)/5-i y/5

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solve for

x, 5 x^{2} - 4 x y + y^{2} = 0

x = \frac{2 y}{5} + i \frac{y}{5}, x = \frac{2 y}{5} - i \frac{y}{5}

Solution steps

5 x^{2} - 4 x y + y^{2} = 0

Write in the standard form

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

5 x^{2} - 4 y x + y^{2} = 0

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

x = \frac{- ( - 4 y ) + 2 i y}{2 \cdot 5} : \frac{2 y}{5} + i \frac{y}{5}

x = \frac{- ( - 4 y ) - 2 i y}{2 \cdot 5} : \frac{2 y}{5} - i \frac{y}{5}

The solutions to the quadratic equation are:

x = \frac{2 y}{5} + i \frac{y}{5}, x = \frac{2 y}{5} - i \frac{y}{5}

9 x^{2} + 4 x - 8 = 0

15 + x (x + 2 x) = 10

2 r^{2} - r + 1 = 0

b^{2} + b - 20 = 0

c^{2} - 6 c - 7 = 0

What are the solutions to the equation solvefor x,5x^2-4xy+y^2=0 ?
The solutions to the equation solvefor x,5x^2-4xy+y^2=0 are x=(2y)/5+i y/5 ,x=(2y)/5-i y/5
Find the zeros of solvefor x,5x^2-4xy+y^2=0
The zeros of solvefor x,5x^2-4xy+y^2=0 are x=(2y)/5+i y/5 ,x=(2y)/5-i y/5