What are the solutions to the equation abx^2+4b^2x+3a^2x+12ab=0 ?

The solutions to the equation abx^2+4b^2x+3a^2x+12ab=0 are x=-(4b)/a ,x=-(3a)/b ;\quad ab\ne 0

Find the zeros of abx^2+4b^2x+3a^2x+12ab=0

The zeros of abx^2+4b^2x+3a^2x+12ab=0 are x=-(4b)/a ,x=-(3a)/b ;\quad ab\ne 0

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ab x^{2} + 4 b^{2} x + 3 a^{2} x + 12 ab = 0

x = - \frac{4 b}{a}, x = - \frac{3 a}{b}; ab \neq = 0

Solution steps

ab x^{2} + 4 b^{2} x + 3 a^{2} x + 12 ab = 0

Write in the standard form

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

ab x^{2} + (4 b^{2} + 3 a^{2}) x + 12 ab = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( 4 b ^{2} + 3 a ^{2} ) \pm ( 4 b ^{2} + 3 a ^{2} ) ^{2} - 4 ab \cdot 12 ab}{2 ab}

Simplify

(4 b^{2} + 3 a^{2})^{2} - 4 ab \cdot 12 ab : (3 a + 2 b) (3 a - 2 b)

x_{1, 2} = \frac{- ( 4 b ^{2} + 3 a ^{2} ) \pm ( 3 a + 2 b ) ( 3 a - 2 b )}{2 ab}; ab \neq = 0

Separate the solutions

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

x = \frac{- ( 4 b ^{2} + 3 a ^{2} ) + ( 3 a + 2 b ) ( 3 a - 2 b )}{2 ab} : - \frac{4 b}{a}

x = \frac{- ( 4 b ^{2} + 3 a ^{2} ) - ( 3 a + 2 b ) ( 3 a - 2 b )}{2 ab} : - \frac{3 a}{b}

The solutions to the quadratic equation are:

x = - \frac{4 b}{a}, x = - \frac{3 a}{b}; ab \neq = 0

3 x^{2} = - 5

b, y = (a - \frac{b}{x})^{2}

2 x^{2} - 7 x = 0

x^{2} - 12 x + 13 = 0

- 0.5 x^{2} + 40 x - 300 = 15000

What are the solutions to the equation abx^2+4b^2x+3a^2x+12ab=0 ?
The solutions to the equation abx^2+4b^2x+3a^2x+12ab=0 are x=-(4b)/a ,x=-(3a)/b ;\quad ab\ne 0
Find the zeros of abx^2+4b^2x+3a^2x+12ab=0
The zeros of abx^2+4b^2x+3a^2x+12ab=0 are x=-(4b)/a ,x=-(3a)/b ;\quad ab\ne 0