What are the solutions to the equation 5x=(2x+1)^2 ?

The solutions to the equation 5x=(2x+1)^2 are x= 1/8+i(sqrt(15))/8 ,x= 1/8-i(sqrt(15))/8

Find the zeros of 5x=(2x+1)^2

The zeros of 5x=(2x+1)^2 are x= 1/8+i(sqrt(15))/8 ,x= 1/8-i(sqrt(15))/8

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5 x = (2 x + 1)^{2}

x = \frac{1}{8} + i \frac{15}{8}, x = \frac{1}{8} - i \frac{15}{8}

Solution steps

5 x = (2 x + 1)^{2}

Expand

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

5 x = 4 x^{2} + 4 x + 1

Switch sides

4 x^{2} + 4 x + 1 = 5 x

Move

5 x

to the left side

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

Solve with the quadratic formula

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

Simplify

(- 1)^{2} - 4 \cdot 4 \cdot 1 : 15 i

x_{1, 2} = \frac{- ( - 1 ) \pm 15 i}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 15 i}{2 \cdot 4}, x_{2} = \frac{- ( - 1 ) - 15 i}{2 \cdot 4}

x = \frac{- ( - 1 ) + 15 i}{2 \cdot 4} : \frac{1}{8} + i \frac{15}{8}

x = \frac{- ( - 1 ) - 15 i}{2 \cdot 4} : \frac{1}{8} - i \frac{15}{8}

The solutions to the quadratic equation are:

x = \frac{1}{8} + i \frac{15}{8}, x = \frac{1}{8} - i \frac{15}{8}

- 4.9 t^{2} + 11 t + 500 = 0

- x^{2} - 4 x + 5

\frac{3}{2} x^{2} + \frac{x}{3} + 1 = 0

4 x^{2} + 16 x - 87 = - 3

0 = 12 x^{2} - 72 x + 81

What are the solutions to the equation 5x=(2x+1)^2 ?
The solutions to the equation 5x=(2x+1)^2 are x= 1/8+i(sqrt(15))/8 ,x= 1/8-i(sqrt(15))/8
Find the zeros of 5x=(2x+1)^2
The zeros of 5x=(2x+1)^2 are x= 1/8+i(sqrt(15))/8 ,x= 1/8-i(sqrt(15))/8