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

The solutions to the equation a^2b^2x^2-(a^2+b^2)x+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-(a^2+b^2)x+1=0

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

(y + \frac{1}{4})^{2} = \frac{6}{9}

9 + 16 = x^{2}

d = d^{1} - d^{2}

0.75 (0.25 - x + x^{2}) = x^{2}

a = 3 x^{2} - 9 x

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