What are the solutions to the equation a^2b^2x^2+b^2x-a^2x-1=0 ?

The solutions to the equation a^2b^2x^2+b^2x-a^2x-1=0 are x= 1/(b^2),x=-1/(a^2);\quad a^2b^2\ne 0

Find the zeros of a^2b^2x^2+b^2x-a^2x-1=0

The zeros of a^2b^2x^2+b^2x-a^2x-1=0 are x= 1/(b^2),x=-1/(a^2);\quad a^2b^2\ne 0

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a^{2} b^{2} x^{2} + b^{2} x - a^{2} x - 1 = 0

x = \frac{1}{b ^{2}}, x = - \frac{1}{a ^{2}}; a^{2} b^{2} \neq = 0

Solution steps

a^{2} b^{2} x^{2} + b^{2} x - a^{2} x - 1 = 0

Write in the standard form

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

a^{2} b^{2} x^{2} + (b^{2} - a^{2}) x - 1 = 0

Solve with the quadratic formula

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

Simplify

(b^{2} - a^{2})^{2} - 4 a^{2} b^{2} (- 1) : a^{2} + b^{2}

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

Separate the solutions

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

x = \frac{- ( b ^{2} - a ^{2} ) + a ^{2} + b ^{2}}{2 a ^{2} b ^{2}} : \frac{1}{b ^{2}}

x = \frac{- ( b ^{2} - a ^{2} ) - ( a ^{2} + b ^{2} )}{2 a ^{2} b ^{2}} : - \frac{1}{a ^{2}}

The solutions to the quadratic equation are:

x = \frac{1}{b ^{2}}, x = - \frac{1}{a ^{2}}; a^{2} b^{2} \neq = 0

100 t^{2} - 192 t + 256 = 0

\frac{a}{b} = k^{2}

x, j = (4 + 3 x) (8 x + s)

x, x - y = 4 y^{3} - 3 x^{2}

(4 a)^{2} + (3 a)^{2} = 12 5^{2}

What are the solutions to the equation a^2b^2x^2+b^2x-a^2x-1=0 ?
The solutions to the equation a^2b^2x^2+b^2x-a^2x-1=0 are x= 1/(b^2),x=-1/(a^2);\quad a^2b^2\ne 0
Find the zeros of a^2b^2x^2+b^2x-a^2x-1=0
The zeros of a^2b^2x^2+b^2x-a^2x-1=0 are x= 1/(b^2),x=-1/(a^2);\quad a^2b^2\ne 0