What are the solutions to the equation ((x-1)^2-3x+1}{15}+\frac{x+1)/5 =0 ?

The solutions to the equation ((x-1)^2-3x+1}{15}+\frac{x+1)/5 =0 are x=1+2i,x=1-2i

Find the zeros of ((x-1)^2-3x+1}{15}+\frac{x+1)/5 =0

The zeros of ((x-1)^2-3x+1}{15}+\frac{x+1)/5 =0 are x=1+2i,x=1-2i

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\frac{( x - 1 ) ^{2} - 3 x + 1}{15} + \frac{x + 1}{5} = 0

x = 1 + 2 i, x = 1 - 2 i

Solution steps

\frac{( x - 1 ) ^{2} - 3 x + 1}{15} + \frac{x + 1}{5} = 0

Find Least Common Multiplier of

15, 5 : 15

Multiply by LCM=

15

\frac{( x - 1 ) ^{2} - 3 x + 1}{15} \cdot 15 + \frac{x + 1}{5} \cdot 15 = 0 \cdot 15

Simplify

(x - 1)^{2} - 3 x + 1 + 3 (x + 1) = 0

Expand

(x - 1)^{2} - 3 x + 1 + 3 (x + 1) : x^{2} - 2 x + 5

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

Simplify

(- 2)^{2} - 4 \cdot 1 \cdot 5 : 4 i

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

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 4 i}{2 \cdot 1}, x_{2} = \frac{- ( - 2 ) - 4 i}{2 \cdot 1}

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

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

The solutions to the quadratic equation are:

x = 1 + 2 i, x = 1 - 2 i

- 3 x + 4 + x \geq - 2 x + 2

4 x - 6 \geq 6 x - 20

7 x + 2 = 16

3 x^{2} - 8 x = 16

\frac{( - 9 ) - 7}{( - 9 ) ^{2} - 49}

What are the solutions to the equation ((x-1)^2-3x+1}{15}+\frac{x+1)/5 =0 ?
The solutions to the equation ((x-1)^2-3x+1}{15}+\frac{x+1)/5 =0 are x=1+2i,x=1-2i
Find the zeros of ((x-1)^2-3x+1}{15}+\frac{x+1)/5 =0
The zeros of ((x-1)^2-3x+1}{15}+\frac{x+1)/5 =0 are x=1+2i,x=1-2i