Algebra Examples

$0 = \frac{1}{3} x^{2} - 3$

Step 1

Move all terms to the left side of the equation and simplify.

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Step 1.1

Simplify the right side.

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Step 1.1.1

Combine $\frac{1}{3}$ and $x^{2}$ .

$0 = \frac{x^{2}}{3} - 3$

Step 1.2

Move all the expressions to the left side of the equation.

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Step 1.2.1

Subtract $\frac{x^{2}}{3}$ from both sides of the equation.

$- \frac{x^{2}}{3} = - 3$

Step 1.2.2

Add $3$ to both sides of the equation.

$- \frac{x^{2}}{3} + 3 = 0$

Step 2

Multiply through by the least common denominator $3$ , then simplify.

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Step 2.1

Apply the distributive property.

$3 (- \frac{x^{2}}{3}) + 3 \cdot 3 = 0$

Step 2.2

Cancel the common factor of $3$ .

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Step 2.2.1

Move the leading negative in $- \frac{x^{2}}{3}$ into the numerator.

$3 (\frac{- x^{2}}{3}) + 3 \cdot 3 = 0$

Step 2.2.2

Cancel the common factor.

$3 (\frac{- x^{2}}{3}) + 3 \cdot 3 = 0$

Step 2.2.3

Rewrite the expression.

$- x^{2} + 3 \cdot 3 = 0$

Step 2.3

Multiply $3$ by $3$ .

$- x^{2} + 9 = 0$

Step 3

Use the quadratic formula to find the solutions.

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

Step 4

Substitute the values $a = - 1$ , $b = 0$ , and $c = 9$ into the quadratic formula and solve for $x$ .

$\frac{0 \pm \sqrt{0^{2} - 4 \cdot (- 1 \cdot 9)}}{2 \cdot - 1}$

Step 5

Simplify.

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Step 5.1

Simplify the numerator.

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Step 5.1.1

Raising $0$ to any positive power yields $0$ .

$x = \frac{0 \pm \sqrt{0 - 4 \cdot - 1 \cdot 9}}{2 \cdot - 1}$

Step 5.1.2

Multiply $- 4 \cdot - 1 \cdot 9$ .

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Step 5.1.2.1

Multiply $- 4$ by $- 1$ .

$x = \frac{0 \pm \sqrt{0 + 4 \cdot 9}}{2 \cdot - 1}$

Step 5.1.2.2

Multiply $4$ by $9$ .

$x = \frac{0 \pm \sqrt{0 + 36}}{2 \cdot - 1}$

Step 5.1.3

Add $0$ and $36$ .

$x = \frac{0 \pm \sqrt{36}}{2 \cdot - 1}$

Step 5.1.4

Rewrite $36$ as $6^{2}$ .

$x = \frac{0 \pm \sqrt{6^{2}}}{2 \cdot - 1}$

Step 5.1.5

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

$x = \frac{0 \pm 6}{2 \cdot - 1}$

Step 5.2

Multiply $2$ by $- 1$ .

$x = \frac{0 \pm 6}{- 2}$

Step 5.3

Simplify $\frac{0 \pm 6}{- 2}$ .

$x = \pm 3$

Step 6

The final answer is the combination of both solutions.

$x = 3, - 3$