What are the solutions to the equation x^2-n^2x+2n=0 ?

The solutions to the equation x^2-n^2x+2n=0 are x=(n^2+sqrt(n^4-8n))/2 ,x=(n^2-sqrt(n^4-8n))/2

Find the zeros of x^2-n^2x+2n=0

The zeros of x^2-n^2x+2n=0 are x=(n^2+sqrt(n^4-8n))/2 ,x=(n^2-sqrt(n^4-8n))/2

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

x = \frac{n ^{2} + n ^{4} - 8 n}{2}, x = \frac{n ^{2} - n ^{4} - 8 n}{2}

Solution steps

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

Solve with the quadratic formula

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

Simplify

(- n^{2})^{2} - 4 \cdot 1 \cdot 2 n : n^{4} - 8 n

x_{1, 2} = \frac{- ( - n ^{2} ) \pm n ^{4} - 8 n}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - n ^{2} ) + n ^{4} - 8 n}{2 \cdot 1}, x_{2} = \frac{- ( - n ^{2} ) - n ^{4} - 8 n}{2 \cdot 1}

x = \frac{- ( - n ^{2} ) + n ^{4} - 8 n}{2 \cdot 1} : \frac{n ^{2} + n ^{4} - 8 n}{2}

x = \frac{- ( - n ^{2} ) - n ^{4} - 8 n}{2 \cdot 1} : \frac{n ^{2} - n ^{4} - 8 n}{2}

The solutions to the quadratic equation are:

x = \frac{n ^{2} + n ^{4} - 8 n}{2}, x = \frac{n ^{2} - n ^{4} - 8 n}{2}

4 x^{2} - 2 x + 5 = 0

(x - 2)^{2} + (x + 4)^{2} = 1

2 x^{2} + 8 x = 10

3 x^{2} - 15 = - 378

t^{2} + 4 t + 3 = 0

What are the solutions to the equation x^2-n^2x+2n=0 ?
The solutions to the equation x^2-n^2x+2n=0 are x=(n^2+sqrt(n^4-8n))/2 ,x=(n^2-sqrt(n^4-8n))/2
Find the zeros of x^2-n^2x+2n=0
The zeros of x^2-n^2x+2n=0 are x=(n^2+sqrt(n^4-8n))/2 ,x=(n^2-sqrt(n^4-8n))/2