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

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

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

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

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

x = - 1 + 5, x = - 1 - 5

Solution steps

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

Expand

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

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

Switch sides

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

Move

x^{2}

to the left side

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 1 \cdot (- 4) = 25

x_{1, 2} = \frac{- 2 \pm 2 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 2 5}{2 \cdot 1}, x_{2} = \frac{- 2 - 2 5}{2 \cdot 1}

x = \frac{- 2 + 2 5}{2 \cdot 1} : - 1 + 5

x = \frac{- 2 - 2 5}{2 \cdot 1} : - 1 - 5

The solutions to the quadratic equation are:

x = - 1 + 5, x = - 1 - 5

- 3 (4 x - 5) = - 33

11 + 7 x \geq - 17

2 (- \frac{1}{3})^{0}

\frac{63 x y ^{3}}{7 y}

- x^{2} + x

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