Solve by Completing the Square 6x^2+5x=4

$6 x^{2} + 5 x = 4$

Step 1

Divide each term in $6 x^{2} + 5 x = 4$ by $6$ and simplify.

Divide each term in $6 x^{2} + 5 x = 4$ by $6$ .

$\frac{6 x^{2}}{6} + \frac{5 x}{6} = \frac{4}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 x^{2}}{6} + \frac{5 x}{6} = \frac{4}{6}$

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

$x^{2} + \frac{5 x}{6} = \frac{4}{6}$

Simplify the right side.

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

Factor $2$ out of $4$ .

$x^{2} + \frac{5 x}{6} = \frac{2 (2)}{6}$

Cancel the common factors.

Factor $2$ out of $6$ .

$x^{2} + \frac{5 x}{6} = \frac{2 \cdot 2}{2 \cdot 3}$

Cancel the common factor.

$x^{2} + \frac{5 x}{6} = \frac{2 \cdot 2}{2 \cdot 3}$

Rewrite the expression.

$x^{2} + \frac{5 x}{6} = \frac{2}{3}$

Step 2

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

${(\frac{b}{2})}^{2} = {(\frac{5}{12})}^{2}$

Step 3

Add the term to each side of the equation.

$x^{2} + \frac{5 x}{6} + {(\frac{5}{12})}^{2} = \frac{2}{3} + {(\frac{5}{12})}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

Simplify each term.

Apply the product rule to $\frac{5}{12}$ .

$x^{2} + \frac{5 x}{6} + \frac{5^{2}}{12^{2}} = \frac{2}{3} + {(\frac{5}{12})}^{2}$

Raise $5$ to the power of $2$ .

$x^{2} + \frac{5 x}{6} + \frac{25}{12^{2}} = \frac{2}{3} + {(\frac{5}{12})}^{2}$

Raise $12$ to the power of $2$ .

$x^{2} + \frac{5 x}{6} + \frac{25}{144} = \frac{2}{3} + {(\frac{5}{12})}^{2}$

Simplify the right side.

Simplify $\frac{2}{3} + {(\frac{5}{12})}^{2}$ .

Simplify each term.

Apply the product rule to $\frac{5}{12}$ .

$x^{2} + \frac{5 x}{6} + \frac{25}{144} = \frac{2}{3} + \frac{5^{2}}{12^{2}}$

Raise $5$ to the power of $2$ .

$x^{2} + \frac{5 x}{6} + \frac{25}{144} = \frac{2}{3} + \frac{25}{12^{2}}$

Raise $12$ to the power of $2$ .

$x^{2} + \frac{5 x}{6} + \frac{25}{144} = \frac{2}{3} + \frac{25}{144}$

To write $\frac{2}{3}$ as a fraction with a common denominator, multiply by $\frac{48}{48}$ .

$x^{2} + \frac{5 x}{6} + \frac{25}{144} = \frac{2}{3} \cdot \frac{48}{48} + \frac{25}{144}$

Write each expression with a common denominator of $144$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{2}{3}$ by $\frac{48}{48}$ .

$x^{2} + \frac{5 x}{6} + \frac{25}{144} = \frac{2 \cdot 48}{3 \cdot 48} + \frac{25}{144}$

Multiply $3$ by $48$ .

$x^{2} + \frac{5 x}{6} + \frac{25}{144} = \frac{2 \cdot 48}{144} + \frac{25}{144}$

Combine the numerators over the common denominator.

$x^{2} + \frac{5 x}{6} + \frac{25}{144} = \frac{2 \cdot 48 + 25}{144}$

Simplify the numerator.

Multiply $2$ by $48$ .

$x^{2} + \frac{5 x}{6} + \frac{25}{144} = \frac{96 + 25}{144}$

Add $96$ and $25$ .

$x^{2} + \frac{5 x}{6} + \frac{25}{144} = \frac{121}{144}$

Step 5

Factor the perfect trinomial square into ${(x + \frac{5}{12})}^{2}$ .

${(x + \frac{5}{12})}^{2} = \frac{121}{144}$

Step 6

Solve the equation for $x$ .

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

$x + \frac{5}{12} = \pm \sqrt{\frac{121}{144}}$

Simplify $\pm \sqrt{\frac{121}{144}}$ .

Rewrite $\sqrt{\frac{121}{144}}$ as $\frac{\sqrt{121}}{\sqrt{144}}$ .

$x + \frac{5}{12} = \pm \frac{\sqrt{121}}{\sqrt{144}}$

Simplify the numerator.

Rewrite $121$ as $11^{2}$ .

$x + \frac{5}{12} = \pm \frac{\sqrt{11^{2}}}{\sqrt{144}}$

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

$x + \frac{5}{12} = \pm \frac{11}{\sqrt{144}}$

Simplify the denominator.

Rewrite $144$ as $12^{2}$ .

$x + \frac{5}{12} = \pm \frac{11}{\sqrt{12^{2}}}$

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

$x + \frac{5}{12} = \pm \frac{11}{12}$

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.

$x + \frac{5}{12} = \frac{11}{12}$

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

Subtract $\frac{5}{12}$ from both sides of the equation.

$x = \frac{11}{12} - \frac{5}{12}$

Combine the numerators over the common denominator.

$x = \frac{11 - 5}{12}$

Subtract $5$ from $11$ .

$x = \frac{6}{12}$

Cancel the common factor of $6$ and $12$ .

Factor $6$ out of $6$ .

$x = \frac{6 (1)}{12}$

Cancel the common factors.

Factor $6$ out of $12$ .

$x = \frac{6 \cdot 1}{6 \cdot 2}$

Cancel the common factor.

$x = \frac{6 \cdot 1}{6 \cdot 2}$

Rewrite the expression.

$x = \frac{1}{2}$

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

$x + \frac{5}{12} = - \frac{11}{12}$

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

Subtract $\frac{5}{12}$ from both sides of the equation.

$x = - \frac{11}{12} - \frac{5}{12}$

Combine the numerators over the common denominator.

$x = \frac{- 11 - 5}{12}$

Subtract $5$ from $- 11$ .

$x = \frac{- 16}{12}$

Cancel the common factor of $- 16$ and $12$ .

Factor $4$ out of $- 16$ .

$x = \frac{4 (- 4)}{12}$

Cancel the common factors.

Factor $4$ out of $12$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

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

$x = \frac{1}{2}, - \frac{4}{3}$

Solve by Completing the Square 6x^2+5x=4

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