What are the solutions to the equation x^2+(n+1)x+n=0 ?

The solutions to the equation x^2+(n+1)x+n=0 are x=-1,x=-n

Find the zeros of x^2+(n+1)x+n=0

The zeros of x^2+(n+1)x+n=0 are x=-1,x=-n

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x^{2} + (n + 1) x + n = 0

x = - 1, x = - n

Solution steps

x^{2} + (n + 1) x + n = 0

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 1, x = - n

5 x^{2} + 9 x + 1

x^{2} - 8 x + 12 = 0

a, y = a^{2}

2 m^{2} - 5 m - 3 = 0

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

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