What are the solutions to the equation solvefor x,x^2+2mx+1=0 ?

The solutions to the equation solvefor x,x^2+2mx+1=0 are x=-m+sqrt(m^2-1),x=-m-sqrt(m^2-1)

Find the zeros of solvefor x,x^2+2mx+1=0

The zeros of solvefor x,x^2+2mx+1=0 are x=-m+sqrt(m^2-1),x=-m-sqrt(m^2-1)

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x, x^{2} + 2 m x + 1 = 0

x = - m + m^{2} - 1, x = - m - m^{2} - 1

Solution steps

x^{2} + 2 m x + 1 = 0

Solve with the quadratic formula

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

Simplify

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

x_{1, 2} = \frac{- 2 m \pm 2 m ^{2} - 1}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 m + 2 m ^{2} - 1}{2 \cdot 1}, x_{2} = \frac{- 2 m - 2 m ^{2} - 1}{2 \cdot 1}

x = \frac{- 2 m + 2 m ^{2} - 1}{2 \cdot 1} : - m + m^{2} - 1

x = \frac{- 2 m - 2 m ^{2} - 1}{2 \cdot 1} : - m - m^{2} - 1

The solutions to the quadratic equation are:

x = - m + m^{2} - 1, x = - m - m^{2} - 1

x^{2} - \frac{9}{4} = 0

2 x^{2} - \frac{14 10}{5} x

x, x^{2} + y^{2} + z^{2} - 2 x - 4 y + 4 z = 0

- 2 x^{2} + 2 x + 8 = 0

x, x^{2} y + x y^{2} = 4

What are the solutions to the equation solvefor x,x^2+2mx+1=0 ?
The solutions to the equation solvefor x,x^2+2mx+1=0 are x=-m+sqrt(m^2-1),x=-m-sqrt(m^2-1)
Find the zeros of solvefor x,x^2+2mx+1=0
The zeros of solvefor x,x^2+2mx+1=0 are x=-m+sqrt(m^2-1),x=-m-sqrt(m^2-1)