What are the solutions to the equation solvefor x,a(1-x/k)x-bxy=0 ?

The solutions to the equation solvefor x,a(1-x/k)x-bxy=0 are x=0,x=(k(-by+a))/a

Find the zeros of solvefor x,a(1-x/k)x-bxy=0

The zeros of solvefor x,a(1-x/k)x-bxy=0 are x=0,x=(k(-by+a))/a

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solve for

x, a (1 - \frac{x}{k}) x - b x y = 0

x = 0, x = \frac{k ( - b y + a )}{a}

Solution steps

a (1 - \frac{x}{k}) x - b x y = 0

Expand

a (1 - \frac{x}{k}) x - b x y : a x - \frac{a x ^{2}}{k} - b x y

a x - \frac{a x ^{2}}{k} - b x y = 0

Write in the standard form

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

- \frac{a x ^{2}}{k} + (a - b y) x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( a - b y ) \pm ( a - b y ) ^{2} - 4 ( - \frac{a}{k} ) \cdot 0}{2 ( - \frac{a}{k} )}

Simplify

(a - b y)^{2} - 4 (- \frac{a}{k}) \cdot 0 : - b y + a

x_{1, 2} = \frac{- ( a - b y ) \pm ( - b y + a )}{2 ( - \frac{a}{k} )}

Separate the solutions

x_{1} = \frac{- ( a - b y ) - b y + a}{2 ( - \frac{a}{k} )}, x_{2} = \frac{- ( a - b y ) - ( - b y + a )}{2 ( - \frac{a}{k} )}

x = \frac{- ( a - b y ) - b y + a}{2 ( - \frac{a}{k} )} : 0

x = \frac{- ( a - b y ) - ( - b y + a )}{2 ( - \frac{a}{k} )} : \frac{k ( - b y + a )}{a}

The solutions to the quadratic equation are:

x = 0, x = \frac{k ( - b y + a )}{a}

x^{2} - 16 + 64 = 0

12 x^{2} + 2 x = 0

0 = 4 x^{2} + 6 x + 2

x \cdot x = x

x, 2 y^{2} + 3 x^{2} = 1

What are the solutions to the equation solvefor x,a(1-x/k)x-bxy=0 ?
The solutions to the equation solvefor x,a(1-x/k)x-bxy=0 are x=0,x=(k(-by+a))/a
Find the zeros of solvefor x,a(1-x/k)x-bxy=0
The zeros of solvefor x,a(1-x/k)x-bxy=0 are x=0,x=(k(-by+a))/a