What are the solutions to the equation (2x-1)(2x+1)=(2x-3)2 ?

The solutions to the equation (2x-1)(2x+1)=(2x-3)2 are x= 1/2+i,x= 1/2-i

Find the zeros of (2x-1)(2x+1)=(2x-3)2

The zeros of (2x-1)(2x+1)=(2x-3)2 are x= 1/2+i,x= 1/2-i

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(2 x - 1) (2 x + 1) = (2 x - 3) 2

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

Solution steps

(2 x - 1) (2 x + 1) = (2 x - 3) \cdot 2

Expand

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

Expand

(2 x - 3) \cdot 2 : 4 x - 6

4 x^{2} - 1 = 4 x - 6

Move

6

to the left side

4 x^{2} + 5 = 4 x

Move

4 x

to the left side

4 x^{2} + 5 - 4 x = 0

Write in the standard form

a x^{2} + b x + c = 0

4 x^{2} - 4 x + 5 = 0

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 8 i}{2 \cdot 4}, x_{2} = \frac{- ( - 4 ) - 8 i}{2 \cdot 4}

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

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

The solutions to the quadratic equation are:

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

x, f^{1} (x) = x^{2} + 2

2 y (y + 5) - 5 = 0

y^{2} + 18 y - 40 = 0

x^{2} + 6 x = 55

k (k - 1) = 12

What are the solutions to the equation (2x-1)(2x+1)=(2x-3)2 ?
The solutions to the equation (2x-1)(2x+1)=(2x-3)2 are x= 1/2+i,x= 1/2-i
Find the zeros of (2x-1)(2x+1)=(2x-3)2
The zeros of (2x-1)(2x+1)=(2x-3)2 are x= 1/2+i,x= 1/2-i