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

The solutions to the equation (2x+1)^2=x(3x+2) are x=-1

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

The zeros of (2x+1)^2=x(3x+2) are x=-1

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

x = - 1

Solution steps

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

Expand

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

Expand

x (3 x + 2) : 3 x^{2} + 2 x

4 x^{2} + 4 x + 1 = 3 x^{2} + 2 x

Move

2 x

to the left side

4 x^{2} + 2 x + 1 = 3 x^{2}

Move

3 x^{2}

to the left side

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 1 \cdot 1 = 0

x_{1, 2} = \frac{- 2 \pm 0}{2 \cdot 1}

x = \frac{- 2}{2 \cdot 1}

\frac{- 2}{2 \cdot 1} = - 1

x = - 1

The solution to the quadratic equation is:

x = - 1

8 x^{2} + 16 x = 42

3 x^{2} + 18 x + 21 = 0

81 = π r^{2}

- 2 x^{2} + 4 x - 20 = 0

x^{2} = 8 x + 1

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