What are the solutions to the equation (2x-1)(8x-4)=-1 ?

The solutions to the equation (2x-1)(8x-4)=-1 are x= 1/2+i 1/4 ,x= 1/2-i 1/4

Find the zeros of (2x-1)(8x-4)=-1

The zeros of (2x-1)(8x-4)=-1 are x= 1/2+i 1/4 ,x= 1/2-i 1/4

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(2 x - 1) (8 x - 4) = - 1

x = \frac{1}{2} + i \frac{1}{4}, x = \frac{1}{2} - i \frac{1}{4}

Solution steps

(2 x - 1) (8 x - 4) = - 1

Expand

(2 x - 1) (8 x - 4) : 16 x^{2} - 16 x + 4

16 x^{2} - 16 x + 4 = - 1

Move

1

to the left side

16 x^{2} - 16 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 16 \cdot 5}{2 \cdot 16}

Simplify

(- 16)^{2} - 4 \cdot 16 \cdot 5 : 8 i

x_{1, 2} = \frac{- ( - 16 ) \pm 8 i}{2 \cdot 16}

Separate the solutions

x_{1} = \frac{- ( - 16 ) + 8 i}{2 \cdot 16}, x_{2} = \frac{- ( - 16 ) - 8 i}{2 \cdot 16}

x = \frac{- ( - 16 ) + 8 i}{2 \cdot 16} : \frac{1}{2} + i \frac{1}{4}

x = \frac{- ( - 16 ) - 8 i}{2 \cdot 16} : \frac{1}{2} - i \frac{1}{4}

The solutions to the quadratic equation are:

x = \frac{1}{2} + i \frac{1}{4}, x = \frac{1}{2} - i \frac{1}{4}

(x + 6) (x + 6) = 0

(x + 2) (x + 5) = (x - 3) (2 x + 4)

7 x^{2} + 3 x + 6 = 8

10 x^{2} + 13 x - 30 = 0

6 h^{2} + 3 h = 0

What are the solutions to the equation (2x-1)(8x-4)=-1 ?
The solutions to the equation (2x-1)(8x-4)=-1 are x= 1/2+i 1/4 ,x= 1/2-i 1/4
Find the zeros of (2x-1)(8x-4)=-1
The zeros of (2x-1)(8x-4)=-1 are x= 1/2+i 1/4 ,x= 1/2-i 1/4