What are the solutions to the equation (sqrt(5)-1)x^2-2x-(sqrt(5)+1)=0 ?

The solutions to the equation (sqrt(5)-1)x^2-2x-(sqrt(5)+1)=0 are x=(3+sqrt(5))/2 ,x=-1

Find the zeros of (sqrt(5)-1)x^2-2x-(sqrt(5)+1)=0

The zeros of (sqrt(5)-1)x^2-2x-(sqrt(5)+1)=0 are x=(3+sqrt(5))/2 ,x=-1

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(5 - 1) x^{2} - 2 x - (5 + 1) = 0

x = \frac{3 + 5}{2}, x = - 1

Decimal Notation

x = 2.61803 \dots, x = - 1

Solution steps

(5 - 1) x^{2} - 2 x - (5 + 1) = 0

Expand

(5 - 1) x^{2} - 2 x - (5 + 1) : 5 x^{2} - x^{2} - 2 x - 5 - 1

5 x^{2} - x^{2} - 2 x - 5 - 1 = 0

Write in the standard form

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

(5 - 1) x^{2} - 2 x - 5 - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( 5 - 1 ) ( - 5 - 1 )}{2 ( 5 - 1 )}

(- 2)^{2} - 4 (5 - 1) (- 5 - 1) = 25

x_{1, 2} = \frac{- ( - 2 ) \pm 2 5}{2 ( 5 - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 5}{2 ( 5 - 1 )}, x_{2} = \frac{- ( - 2 ) - 2 5}{2 ( 5 - 1 )}

x = \frac{- ( - 2 ) + 2 5}{2 ( 5 - 1 )} : \frac{3 + 5}{2}

x = \frac{- ( - 2 ) - 2 5}{2 ( 5 - 1 )} : - 1

The solutions to the quadratic equation are:

x = \frac{3 + 5}{2}, x = - 1

x, h^{- 1} (x) = \frac{1}{2} x^{2}

5 a^{2} - 7 a - 10 = 0

2 x^{2} + 24 x + 32 = 0

x^{2} + 15 x + 54 = 0

x^{2} = 6 x + 1

What are the solutions to the equation (sqrt(5)-1)x^2-2x-(sqrt(5)+1)=0 ?
The solutions to the equation (sqrt(5)-1)x^2-2x-(sqrt(5)+1)=0 are x=(3+sqrt(5))/2 ,x=-1
Find the zeros of (sqrt(5)-1)x^2-2x-(sqrt(5)+1)=0
The zeros of (sqrt(5)-1)x^2-2x-(sqrt(5)+1)=0 are x=(3+sqrt(5))/2 ,x=-1