Solve by Completing the Square 3x^2-2x-5=0

$3 x^{2} - 2 x - 5 = 0$

Step 1

Add $5$ to both sides of the equation.

$3 x^{2} - 2 x = 5$

Step 2

Divide each term in $3 x^{2} - 2 x = 5$ by $3$ and simplify.

Divide each term in $3 x^{2} - 2 x = 5$ by $3$ .

$\frac{3 x^{2}}{3} + \frac{- 2 x}{3} = \frac{5}{3}$

Simplify the left side.

Simplify each term.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x^{2}}{3} + \frac{- 2 x}{3} = \frac{5}{3}$

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

$x^{2} + \frac{- 2 x}{3} = \frac{5}{3}$

Move the negative in front of the fraction.

$x^{2} - \frac{2 x}{3} = \frac{5}{3}$

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 $b$ .

${(\frac{b}{2})}^{2} = {(- \frac{1}{3})}^{2}$

Step 4

Add the term to each side of the equation.

$x^{2} - \frac{2 x}{3} + {(- \frac{1}{3})}^{2} = \frac{5}{3} + {(- \frac{1}{3})}^{2}$

Step 5

Simplify the equation.

Simplify the left side.

Simplify each term.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{1}{3}$ .

$x^{2} - \frac{2 x}{3} + {(- 1)}^{2} {(\frac{1}{3})}^{2} = \frac{5}{3} + {(- \frac{1}{3})}^{2}$

Apply the product rule to $\frac{1}{3}$ .

$x^{2} - \frac{2 x}{3} + {(- 1)}^{2} \frac{1^{2}}{3^{2}} = \frac{5}{3} + {(- \frac{1}{3})}^{2}$

Raise $- 1$ to the power of $2$ .

$x^{2} - \frac{2 x}{3} + 1 \frac{1^{2}}{3^{2}} = \frac{5}{3} + {(- \frac{1}{3})}^{2}$

Multiply $\frac{1^{2}}{3^{2}}$ by $1$ .

$x^{2} - \frac{2 x}{3} + \frac{1^{2}}{3^{2}} = \frac{5}{3} + {(- \frac{1}{3})}^{2}$

One to any power is one.

$x^{2} - \frac{2 x}{3} + \frac{1}{3^{2}} = \frac{5}{3} + {(- \frac{1}{3})}^{2}$

Raise $3$ to the power of $2$ .

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{5}{3} + {(- \frac{1}{3})}^{2}$

Simplify the right side.

Simplify $\frac{5}{3} + {(- \frac{1}{3})}^{2}$ .

Simplify each term.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $- \frac{1}{3}$ .

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{5}{3} + {(- 1)}^{2} {(\frac{1}{3})}^{2}$

Apply the product rule to $\frac{1}{3}$ .

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{5}{3} + {(- 1)}^{2} \frac{1^{2}}{3^{2}}$

Raise $- 1$ to the power of $2$ .

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{5}{3} + 1 \frac{1^{2}}{3^{2}}$

Multiply $\frac{1^{2}}{3^{2}}$ by $1$ .

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{5}{3} + \frac{1^{2}}{3^{2}}$

One to any power is one.

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{5}{3} + \frac{1}{3^{2}}$

Raise $3$ to the power of $2$ .

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{5}{3} + \frac{1}{9}$

To write $\frac{5}{3}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{5}{3} \cdot \frac{3}{3} + \frac{1}{9}$

Write each expression with a common denominator of $9$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{5}{3}$ by $\frac{3}{3}$ .

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{5 \cdot 3}{3 \cdot 3} + \frac{1}{9}$

Multiply $3$ by $3$ .

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{5 \cdot 3}{9} + \frac{1}{9}$

Combine the numerators over the common denominator.

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{5 \cdot 3 + 1}{9}$

Simplify the numerator.

Multiply $5$ by $3$ .

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{15 + 1}{9}$

Add $15$ and $1$ .

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{16}{9}$

Step 6

Factor the perfect trinomial square into ${(x - \frac{1}{3})}^{2}$ .

${(x - \frac{1}{3})}^{2} = \frac{16}{9}$

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.

$x - \frac{1}{3} = \pm \sqrt{\frac{16}{9}}$

Simplify $\pm \sqrt{\frac{16}{9}}$ .

Rewrite $\sqrt{\frac{16}{9}}$ as $\frac{\sqrt{16}}{\sqrt{9}}$ .

$x - \frac{1}{3} = \pm \frac{\sqrt{16}}{\sqrt{9}}$

Simplify the numerator.

Rewrite $16$ as $4^{2}$ .

$x - \frac{1}{3} = \pm \frac{\sqrt{4^{2}}}{\sqrt{9}}$

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

$x - \frac{1}{3} = \pm \frac{4}{\sqrt{9}}$

Simplify the denominator.

Rewrite $9$ as $3^{2}$ .

$x - \frac{1}{3} = \pm \frac{4}{\sqrt{3^{2}}}$

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

$x - \frac{1}{3} = \pm \frac{4}{3}$

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.

$x - \frac{1}{3} = \frac{4}{3}$

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

Add $\frac{1}{3}$ to both sides of the equation.

$x = \frac{4}{3} + \frac{1}{3}$

Combine the numerators over the common denominator.

$x = \frac{4 + 1}{3}$

Add $4$ and $1$ .

$x = \frac{5}{3}$

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

$x - \frac{1}{3} = - \frac{4}{3}$

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

Add $\frac{1}{3}$ to both sides of the equation.

$x = - \frac{4}{3} + \frac{1}{3}$

Combine the numerators over the common denominator.

$x = \frac{- 4 + 1}{3}$

Add $- 4$ and $1$ .

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

Divide $- 3$ by $3$ .

$x = - 1$

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

$x = \frac{5}{3}, - 1$

Solve by Completing the Square 3x^2-2x-5=0

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