Solve by Completing the Square 3x^2+x-2=0

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

Step 1

Add $2$ to both sides of the equation.

$3 x^{2} + x = 2$

Step 2

Divide each term in $3 x^{2} + x = 2$ by $3$ and simplify.

Divide each term in $3 x^{2} + x = 2$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x^{2}$ by $1$ .

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

Step 3

To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of $b$ .

${(\frac{b}{2})}^{2} = {(\frac{1}{6})}^{2}$

Step 4

Add the term to each side of the equation.

$x^{2} + \frac{x}{3} + {(\frac{1}{6})}^{2} = \frac{2}{3} + {(\frac{1}{6})}^{2}$

Step 5

Simplify the equation.

Simplify the left side.

Simplify each term.

Apply the product rule to $\frac{1}{6}$ .

$x^{2} + \frac{x}{3} + \frac{1^{2}}{6^{2}} = \frac{2}{3} + {(\frac{1}{6})}^{2}$

One to any power is one.

$x^{2} + \frac{x}{3} + \frac{1}{6^{2}} = \frac{2}{3} + {(\frac{1}{6})}^{2}$

Raise $6$ to the power of $2$ .

$x^{2} + \frac{x}{3} + \frac{1}{36} = \frac{2}{3} + {(\frac{1}{6})}^{2}$

Simplify the right side.

Simplify $\frac{2}{3} + {(\frac{1}{6})}^{2}$ .

Simplify each term.

Apply the product rule to $\frac{1}{6}$ .

$x^{2} + \frac{x}{3} + \frac{1}{36} = \frac{2}{3} + \frac{1^{2}}{6^{2}}$

One to any power is one.

$x^{2} + \frac{x}{3} + \frac{1}{36} = \frac{2}{3} + \frac{1}{6^{2}}$

Raise $6$ to the power of $2$ .

$x^{2} + \frac{x}{3} + \frac{1}{36} = \frac{2}{3} + \frac{1}{36}$

To write $\frac{2}{3}$ as a fraction with a common denominator, multiply by $\frac{12}{12}$ .

$x^{2} + \frac{x}{3} + \frac{1}{36} = \frac{2}{3} \cdot \frac{12}{12} + \frac{1}{36}$

Write each expression with a common denominator of $36$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{2}{3}$ by $\frac{12}{12}$ .

$x^{2} + \frac{x}{3} + \frac{1}{36} = \frac{2 \cdot 12}{3 \cdot 12} + \frac{1}{36}$

Multiply $3$ by $12$ .

$x^{2} + \frac{x}{3} + \frac{1}{36} = \frac{2 \cdot 12}{36} + \frac{1}{36}$

Combine the numerators over the common denominator.

$x^{2} + \frac{x}{3} + \frac{1}{36} = \frac{2 \cdot 12 + 1}{36}$

Simplify the numerator.

Multiply $2$ by $12$ .

$x^{2} + \frac{x}{3} + \frac{1}{36} = \frac{24 + 1}{36}$

Add $24$ and $1$ .

$x^{2} + \frac{x}{3} + \frac{1}{36} = \frac{25}{36}$

Step 6

Factor the perfect trinomial square into ${(x + \frac{1}{6})}^{2}$ .

${(x + \frac{1}{6})}^{2} = \frac{25}{36}$

Step 7

Solve the equation for $x$ .

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x + \frac{1}{6} = \pm \sqrt{\frac{25}{36}}$

Simplify $\pm \sqrt{\frac{25}{36}}$ .

Rewrite $\sqrt{\frac{25}{36}}$ as $\frac{\sqrt{25}}{\sqrt{36}}$ .

$x + \frac{1}{6} = \pm \frac{\sqrt{25}}{\sqrt{36}}$

Simplify the numerator.

Rewrite $25$ as $5^{2}$ .

$x + \frac{1}{6} = \pm \frac{\sqrt{5^{2}}}{\sqrt{36}}$

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

$x + \frac{1}{6} = \pm \frac{5}{\sqrt{36}}$

Simplify the denominator.

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

$x + \frac{1}{6} = \pm \frac{5}{\sqrt{6^{2}}}$

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

$x + \frac{1}{6} = \pm \frac{5}{6}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x + \frac{1}{6} = \frac{5}{6}$

Move all terms not containing $x$ to the right side of the equation.

Subtract $\frac{1}{6}$ from both sides of the equation.

$x = \frac{5}{6} - \frac{1}{6}$

Combine the numerators over the common denominator.

$x = \frac{5 - 1}{6}$

Subtract $1$ from $5$ .

$x = \frac{4}{6}$

Cancel the common factor of $4$ and $6$ .

Factor $2$ out of $4$ .

$x = \frac{2 (2)}{6}$

Cancel the common factors.

Factor $2$ out of $6$ .

$x = \frac{2 \cdot 2}{2 \cdot 3}$

Cancel the common factor.

$x = \frac{2 \cdot 2}{2 \cdot 3}$

Rewrite the expression.

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

Next, use the negative value of the $\pm$ to find the second solution.

$x + \frac{1}{6} = - \frac{5}{6}$

Move all terms not containing $x$ to the right side of the equation.

Subtract $\frac{1}{6}$ from both sides of the equation.

$x = - \frac{5}{6} - \frac{1}{6}$

Combine the numerators over the common denominator.

$x = \frac{- 5 - 1}{6}$

Subtract $1$ from $- 5$ .

$x = \frac{- 6}{6}$

Divide $- 6$ by $6$ .

$x = - 1$

The complete solution is the result of both the positive and negative portions of the solution.

$x = \frac{2}{3}, - 1$

Solve by Completing the Square 3x^2+x-2=0

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