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

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

Find the zeros of x^2+(a/(a+b)+(a+b)/a)x+1=0

The zeros of x^2+(a/(a+b)+(a+b)/a)x+1=0 are

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

x = \frac{- 2 ( a ^{2} + ab ) - b ^{2} + \frac{b ^{2} ( 4 a ^{5} + 16 a ^{4} b + 25 a ^{3} b ^{2} + 19 a ^{2} b ^{3} + 7 a b ^{4} + b ^{5} )}{( a + b ) ^{3} ( a ^{2} + ab ) ^{2}} ( a ^{2} + ab )}{2 ( a ^{2} + ab )}, x = \frac{- 2 ( a ^{2} + ab ) - b ^{2} - \frac{b ^{2} ( 4 a ^{5} + 16 a ^{4} b + 25 a ^{3} b ^{2} + 19 a ^{2} b ^{3} + 7 a b ^{4} + b ^{5} )}{( a + b ) ^{3} ( a ^{2} + ab ) ^{2}} ( a ^{2} + ab )}{2 ( a ^{2} + ab )}

Solution steps

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

Expand

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

x^{2} + \frac{2 a ^{2} x + 2 ab x + b ^{2} x}{a ^{2} + ab} + 1 = 0

Write in the standard form

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

x^{2} + (2 + \frac{b ^{2}}{a ^{2} + ab}) x + 1 = 0

Solve with the quadratic formula

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

Simplify

(2 + \frac{b ^{2}}{a ^{2} + ab})^{2} - 4 \cdot 1 \cdot 1 : \frac{b ^{2} ( 4 a ^{5} + 16 a ^{4} b + 25 a ^{3} b ^{2} + 19 a ^{2} b ^{3} + 7 a b ^{4} + b ^{5} )}{( a + b ) ^{3} ( a ^{2} + ab ) ^{2}}

x_{1, 2} = \frac{- ( 2 + \frac{b ^{2}}{a ^{2} + ab} ) \pm \frac{b ^{2} ( 4 a ^{5} + 16 a ^{4} b + 25 a ^{3} b ^{2} + 19 a ^{2} b ^{3} + 7 a b ^{4} + b ^{5} )}{( a + b ) ^{3} ( a ^{2} + ab ) ^{2}}}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( 2 + \frac{b ^{2}}{a ^{2} + ab} ) + \frac{b ^{2} ( 4 a ^{5} + 16 a ^{4} b + 25 a ^{3} b ^{2} + 19 a ^{2} b ^{3} + 7 a b ^{4} + b ^{5} )}{( a + b ) ^{3} ( a ^{2} + ab ) ^{2}}}{2 \cdot 1}, x_{2} = \frac{- ( 2 + \frac{b ^{2}}{a ^{2} + ab} ) - \frac{b ^{2} ( 4 a ^{5} + 16 a ^{4} b + 25 a ^{3} b ^{2} + 19 a ^{2} b ^{3} + 7 a b ^{4} + b ^{5} )}{( a + b ) ^{3} ( a ^{2} + ab ) ^{2}}}{2 \cdot 1}

x = \frac{- ( 2 + \frac{b ^{2}}{a ^{2} + ab} ) + \frac{b ^{2} ( 4 a ^{5} + 16 a ^{4} b + 25 a ^{3} b ^{2} + 19 a ^{2} b ^{3} + 7 a b ^{4} + b ^{5} )}{( a + b ) ^{3} ( a ^{2} + ab ) ^{2}}}{2 \cdot 1} : \frac{- 2 ( a ^{2} + ab ) - b ^{2} + \frac{b ^{2} ( 4 a ^{5} + 16 a ^{4} b + 25 a ^{3} b ^{2} + 19 a ^{2} b ^{3} + 7 a b ^{4} + b ^{5} )}{( a + b ) ^{3} ( a ^{2} + ab ) ^{2}} ( a ^{2} + ab )}{2 ( a ^{2} + ab )}

x = \frac{- ( 2 + \frac{b ^{2}}{a ^{2} + ab} ) - \frac{b ^{2} ( 4 a ^{5} + 16 a ^{4} b + 25 a ^{3} b ^{2} + 19 a ^{2} b ^{3} + 7 a b ^{4} + b ^{5} )}{( a + b ) ^{3} ( a ^{2} + ab ) ^{2}}}{2 \cdot 1} : \frac{- 2 ( a ^{2} + ab ) - b ^{2} - \frac{b ^{2} ( 4 a ^{5} + 16 a ^{4} b + 25 a ^{3} b ^{2} + 19 a ^{2} b ^{3} + 7 a b ^{4} + b ^{5} )}{( a + b ) ^{3} ( a ^{2} + ab ) ^{2}} ( a ^{2} + ab )}{2 ( a ^{2} + ab )}

The solutions to the quadratic equation are:

x = \frac{- 2 ( a ^{2} + ab ) - b ^{2} + \frac{b ^{2} ( 4 a ^{5} + 16 a ^{4} b + 25 a ^{3} b ^{2} + 19 a ^{2} b ^{3} + 7 a b ^{4} + b ^{5} )}{( a + b ) ^{3} ( a ^{2} + ab ) ^{2}} ( a ^{2} + ab )}{2 ( a ^{2} + ab )}, x = \frac{- 2 ( a ^{2} + ab ) - b ^{2} - \frac{b ^{2} ( 4 a ^{5} + 16 a ^{4} b + 25 a ^{3} b ^{2} + 19 a ^{2} b ^{3} + 7 a b ^{4} + b ^{5} )}{( a + b ) ^{3} ( a ^{2} + ab ) ^{2}} ( a ^{2} + ab )}{2 ( a ^{2} + ab )}

3 x^{2} - x - 24 = 0

6 x^{2} - 8 x + 5 = 0

2 x^{2} - 8 x - 16 = 0

x^{2} + 12 x + 50 = 15

x + x^{2} - 1 = 0

What are the solutions to the equation x^2+(a/(a+b)+(a+b)/a)x+1=0 ?
The solutions to the equation x^2+(a/(a+b)+(a+b)/a)x+1=0 are
Find the zeros of x^2+(a/(a+b)+(a+b)/a)x+1=0
The zeros of x^2+(a/(a+b)+(a+b)/a)x+1=0 are