What are the solutions to the equation solvefor x,64x^2-48xy+9y^2=0 ?

The solutions to the equation solvefor x,64x^2-48xy+9y^2=0 are x=(3y)/8

Find the zeros of solvefor x,64x^2-48xy+9y^2=0

The zeros of solvefor x,64x^2-48xy+9y^2=0 are x=(3y)/8

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solve for

x, 64 x^{2} - 48 x y + 9 y^{2} = 0

x = \frac{3 y}{8}

Solution steps

64 x^{2} - 48 x y + 9 y^{2} = 0

Write in the standard form

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

64 x^{2} - 48 y x + 9 y^{2} = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 48 y ) \pm ( - 48 y ) ^{2} - 4 \cdot 64 \cdot 9 y ^{2}}{2 \cdot 64}

(- 48 y)^{2} - 4 \cdot 64 \cdot 9 y^{2} = 0

x_{1, 2} = \frac{- ( - 48 y ) \pm 0}{2 \cdot 64}

x = \frac{- ( - 48 y )}{2 \cdot 64}

Simplify

\frac{- ( - 48 y )}{2 \cdot 64} : \frac{3 y}{8}

x = \frac{3 y}{8}

The solution to the quadratic equation is:

x = \frac{3 y}{8}

x^{2} (- 2) = 16

x^{2} + x + 32 = 0

y, (x + y)^{2} = x

r^{2} - 8 r - 22 = 6

78 = π r^{2} (\frac{101}{360})

What are the solutions to the equation solvefor x,64x^2-48xy+9y^2=0 ?
The solutions to the equation solvefor x,64x^2-48xy+9y^2=0 are x=(3y)/8
Find the zeros of solvefor x,64x^2-48xy+9y^2=0
The zeros of solvefor x,64x^2-48xy+9y^2=0 are x=(3y)/8