What are the solutions to the equation (n-1)^2=2n ?

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

Find the zeros of (n-1)^2=2n

The zeros of (n-1)^2=2n are n=2+sqrt(3),n=2-sqrt(3)

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(n - 1)^{2} = 2 n

n = 2 + 3, n = 2 - 3

Decimal Notation

n = 3.73205 \dots, n = 0.26794 \dots

Solution steps

(n - 1)^{2} = 2 n

Expand

(n - 1)^{2} : n^{2} - 2 n + 1

n^{2} - 2 n + 1 = 2 n

Move

2 n

to the left side

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

n = 2 + 3, n = 2 - 3

2 x^{2} - 3 x - 9 > 0

\frac{1}{4} - \frac{2}{7}

2 x + x = 63

- x^{2} + 3 x - 2 \geq 0

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