What are the solutions to the equation x^2-3x+1=0,2<= x<= 3 ?

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

Find the zeros of x^2-3x+1=0,2<= x<= 3

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

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

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

x^{2} - 3 x + 1 = 0, 2 \leq x \leq 3

x = \frac{3 + 5}{2}

Decimal Notation

x = 2.61803 \dots

Solution steps

x^{2} - 3 x + 1 = 0

Solve with the quadratic formula

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

(- 3)^{2} - 4 \cdot 1 \cdot 1 = 5

x_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 5}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 5}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

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

Check constrains

x = \frac{3 + 5}{2}

2 x^{2} + 11 x - 138 = 0

x^{2} - \frac{1}{3} x = 0

p^{2} - 20 p + 100 = 0

4 r^{2} + 8 r + 4 = 0

(6 X - 4) (3 X + 5) = (4 X - 9) (3 X + 5)

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