What are the solutions to the equation (x)=7x^2-x+1 ?

The solutions to the equation (x)=7x^2-x+1 are x= 1/7+i(sqrt(6))/7 ,x= 1/7-i(sqrt(6))/7

Find the zeros of (x)=7x^2-x+1

The zeros of (x)=7x^2-x+1 are x= 1/7+i(sqrt(6))/7 ,x= 1/7-i(sqrt(6))/7

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

x = \frac{1}{7} + i \frac{6}{7}, x = \frac{1}{7} - i \frac{6}{7}

Solution steps

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

Refine

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

Switch sides

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

Move

x

to the left side

7 x^{2} - 2 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 7 \cdot 1}{2 \cdot 7}

Simplify

(- 2)^{2} - 4 \cdot 7 \cdot 1 : 26 i

x_{1, 2} = \frac{- ( - 2 ) \pm 2 6 i}{2 \cdot 7}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 6 i}{2 \cdot 7}, x_{2} = \frac{- ( - 2 ) - 2 6 i}{2 \cdot 7}

x = \frac{- ( - 2 ) + 2 6 i}{2 \cdot 7} : \frac{1}{7} + i \frac{6}{7}

x = \frac{- ( - 2 ) - 2 6 i}{2 \cdot 7} : \frac{1}{7} - i \frac{6}{7}

The solutions to the quadratic equation are:

x = \frac{1}{7} + i \frac{6}{7}, x = \frac{1}{7} - i \frac{6}{7}

x^{2} + x^{2} = 9 2^{2}

λ^{2} - λ - 1 = 0

(x + 1) (x - 4) = 6

x (2 x) = 1800

(3 x + 2)^{2} + (x + 4)^{2} = (2 x + 6)

What are the solutions to the equation (x)=7x^2-x+1 ?
The solutions to the equation (x)=7x^2-x+1 are x= 1/7+i(sqrt(6))/7 ,x= 1/7-i(sqrt(6))/7
Find the zeros of (x)=7x^2-x+1
The zeros of (x)=7x^2-x+1 are x= 1/7+i(sqrt(6))/7 ,x= 1/7-i(sqrt(6))/7