What are the solutions to the equation u^2-4u+1=0 ?

The solutions to the equation u^2-4u+1=0 are u=2+sqrt(3),u=2-sqrt(3)

Find the zeros of u^2-4u+1=0

The zeros of u^2-4u+1=0 are u=2+sqrt(3),u=2-sqrt(3)

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u^{2} - 4 u + 1 = 0

u = 2 + 3, u = 2 - 3

Decimal Notation

u = 3.73205 \dots, u = 0.26794 \dots

Solution steps

u^{2} - 4 u + 1 = 0

Solve with the quadratic formula

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

(- 4)^{2} - 4 \cdot 1 \cdot 1 = 23

u_{1, 2} = \frac{- ( - 4 ) \pm 2 3}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 4 ) + 2 3}{2 \cdot 1}, u_{2} = \frac{- ( - 4 ) - 2 3}{2 \cdot 1}

u = \frac{- ( - 4 ) + 2 3}{2 \cdot 1} : 2 + 3

u = \frac{- ( - 4 ) - 2 3}{2 \cdot 1} : 2 - 3

The solutions to the quadratic equation are:

u = 2 + 3, u = 2 - 3

- 18 x^{2} + 18 x = 0

(x + 1)^{2} = 56

y^{2} - 6 y - 15 = 0

3 x^{2} - 42 x = - 120

x^{2} - 2 x - 19 = 0

What are the solutions to the equation u^2-4u+1=0 ?
The solutions to the equation u^2-4u+1=0 are u=2+sqrt(3),u=2-sqrt(3)
Find the zeros of u^2-4u+1=0
The zeros of u^2-4u+1=0 are u=2+sqrt(3),u=2-sqrt(3)