Solve for x 1/2kx^2=mgx

$\frac{1}{2} k x^{2} = mg x$

Multiply $\frac{1}{2} (k x^{2})$ .

Combine $k$ and $\frac{1}{2}$ .

$x = \frac{k}{2} x^{2}$

Combine $\frac{k}{2}$ and $x^{2}$ .

$x = \frac{k x^{2}}{2}$

Subtract $\frac{k x^{2}}{2}$ from both sides of the equation.

$x - \frac{k x^{2}}{2} = 0$

Multiply through by the least common denominator $2$ , then simplify.

Apply the distributive property.

$2 x + 2 (- \frac{k x^{2}}{2}) = 0$

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{k x^{2}}{2}$ into the numerator.

$2 x + 2 (\frac{- k x^{2}}{2}) = 0$

Cancel the common factor.

$2 x + 2 (\frac{- k x^{2}}{2}) = 0$

Rewrite the expression.

$2 x - k x^{2} = 0$

Reorder $2 x$ and $- k x^{2}$ .

$- k x^{2} + 2 x = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = - k$ , $b = 2$ , and $c = 0$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (- k \cdot 0)}}{2 (- k)}$

Simplify.

Simplify the numerator.

Rewrite $- 4 k \cdot 0$ as $0^{2}$ .

$x = \frac{- 2 \pm \sqrt{2^{2} - 0^{2}}}{2 (- k)}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2$ and $b = 0$ .

$x = \frac{- 2 \pm \sqrt{(2 + 0) (2 + 0)}}{2 (- k)}$

Simplify.

Add $2$ and $0$ .

$x = \frac{- 2 \pm \sqrt{2 (2 + 0)}}{2 (- k)}$

Add $2$ and $0$ .

$x = \frac{- 2 \pm \sqrt{2 \cdot 2}}{2 (- k)}$

Multiply $2$ by $2$ .

$x = \frac{- 2 \pm \sqrt{4}}{2 (- k)}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{- 2 \pm \sqrt{2^{2}}}{2 (- k)}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \frac{- 2 \pm 2}{2 (- k)}$

Multiply $- 1$ by $2$ .

$x = \frac{- 2 \pm 2}{- 2 k}$

Simplify $\frac{- 2 \pm 2}{- 2 k}$ .

$x = \frac{1 \pm 1}{k}$

The final answer is the combination of both solutions.

$x = \frac{2}{k}$

$x = 0$

Solve for x 1/2kx^2=mgx

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