What are the solutions to the equation roots x^2-8x+41 ?

The solutions to the equation roots x^2-8x+41 are x=4+5i,x=4-5i

Find the zeros of roots x^2-8x+41

The zeros of roots x^2-8x+41 are x=4+5i,x=4-5i

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roots

x^{2} - 8 x + 41

x = 4 + 5 i, x = 4 - 5 i

Solution steps

x^{2} - 8 x + 41

The roots are the intercepts with the x-axis

(y = 0)

x^{2} - 8 x + 41 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 1 \cdot 41}{2 \cdot 1}

Simplify

(- 8)^{2} - 4 \cdot 1 \cdot 41 : 10 i

x_{1, 2} = \frac{- ( - 8 ) \pm 10 i}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 10 i}{2 \cdot 1}, x_{2} = \frac{- ( - 8 ) - 10 i}{2 \cdot 1}

x = \frac{- ( - 8 ) + 10 i}{2 \cdot 1} : 4 + 5 i

x = \frac{- ( - 8 ) - 10 i}{2 \cdot 1} : 4 - 5 i

The solutions to the quadratic equation are:

x = 4 + 5 i, x = 4 - 5 i

2 x^{2} + 4 x = - 12

x^{2} - 20 x + 72 = 0

2 x^{2} + 7 x + 2 = - 1

3 x^{2} - 8 x - 1 = - 6 x + 4

25 - 100 x^{2} = 0

What are the solutions to the equation roots x^2-8x+41 ?
The solutions to the equation roots x^2-8x+41 are x=4+5i,x=4-5i
Find the zeros of roots x^2-8x+41
The zeros of roots x^2-8x+41 are x=4+5i,x=4-5i