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

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

Find the zeros of abx^2-(a^2+b^2)x+ab=0

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

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

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

Solution steps

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

5 x^{2} - 7 = 0

5 x^{2} - 11 x = 0

6 x^{2} + 7 x + 9 = 11 x^{2}

2 (x^{2} - 1) + 1 = 1

2 x^{2} - 11 x - 3 = 0

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