What are the solutions to the equation 2(ax^2-b^3)=bx(4b-a) ?

The solutions to the equation 2(ax^2-b^3)=bx(4b-a) are x=(2b^2)/a ,x=-b/2 ;\quad a\ne 0

Find the zeros of 2(ax^2-b^3)=bx(4b-a)

The zeros of 2(ax^2-b^3)=bx(4b-a) are x=(2b^2)/a ,x=-b/2 ;\quad a\ne 0

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2 (a x^{2} - b^{3}) = b x (4 b - a)

x = \frac{2 b ^{2}}{a}, x = - \frac{b}{2}; a \neq = 0

Solution steps

2 (a x^{2} - b^{3}) = b x (4 b - a)

Expand

2 (a x^{2} - b^{3}) : 2 a x^{2} - 2 b^{3}

Expand

b x (4 b - a) : 4 b^{2} x - ab x

2 a x^{2} - 2 b^{3} = 4 b^{2} x - ab x

Move

ab x

to the left side

2 a x^{2} - 2 b^{3} + ab x = 4 b^{2} x

Move

4 b^{2} x

to the left side

2 a x^{2} - 2 b^{3} + ab x - 4 b^{2} x = 0

Write in the standard form

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

2 a x^{2} + (ab - 4 b^{2}) x - 2 b^{3} = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( ab - 4 b ^{2} ) \pm ( ab - 4 b ^{2} ) ^{2} - 4 \cdot 2 a ( - 2 b ^{3} )}{2 \cdot 2 a}

Simplify

(ab - 4 b^{2})^{2} - 4 \cdot 2 a (- 2 b^{3}) : b (a + 4 b)

x_{1, 2} = \frac{- ( ab - 4 b ^{2} ) \pm b ( a + 4 b )}{2 \cdot 2 a}; a \neq = 0

Separate the solutions

x_{1} = \frac{- ( ab - 4 b ^{2} ) + b ( a + 4 b )}{2 \cdot 2 a}, x_{2} = \frac{- ( ab - 4 b ^{2} ) - b ( a + 4 b )}{2 \cdot 2 a}

x = \frac{- ( ab - 4 b ^{2} ) + b ( a + 4 b )}{2 \cdot 2 a} : \frac{2 b ^{2}}{a}

x = \frac{- ( ab - 4 b ^{2} ) - b ( a + 4 b )}{2 \cdot 2 a} : - \frac{b}{2}

The solutions to the quadratic equation are:

x = \frac{2 b ^{2}}{a}, x = - \frac{b}{2}; a \neq = 0

2 x^{2} + x + 2

r^{2} + 5 r - 72 = 0

2 x^{2} - x = 14

(3 x)^{2} = (x + 4)^{2} + (2 x - 3)^{2}

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

What are the solutions to the equation 2(ax^2-b^3)=bx(4b-a) ?
The solutions to the equation 2(ax^2-b^3)=bx(4b-a) are x=(2b^2)/a ,x=-b/2 ;\quad a\ne 0
Find the zeros of 2(ax^2-b^3)=bx(4b-a)
The zeros of 2(ax^2-b^3)=bx(4b-a) are x=(2b^2)/a ,x=-b/2 ;\quad a\ne 0