Solve by Completing the Square 8x^2-2x-15=0

$8 x^{2} - 2 x - 15 = 0$

Step 1

Add $15$ to both sides of the equation.

$8 x^{2} - 2 x = 15$

Step 2

Divide each term in $8 x^{2} - 2 x = 15$ by $8$ and simplify.

Divide each term in $8 x^{2} - 2 x = 15$ by $8$ .

$\frac{8 x^{2}}{8} + \frac{- 2 x}{8} = \frac{15}{8}$

Simplify the left side.

Simplify each term.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 x^{2}}{8} + \frac{- 2 x}{8} = \frac{15}{8}$

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

$x^{2} + \frac{- 2 x}{8} = \frac{15}{8}$

Cancel the common factor of $- 2$ and $8$ .

Factor $2$ out of $- 2 x$ .

$x^{2} + \frac{2 (- x)}{8} = \frac{15}{8}$

Cancel the common factors.

Factor $2$ out of $8$ .

$x^{2} + \frac{2 (- x)}{2 (4)} = \frac{15}{8}$

Cancel the common factor.

$x^{2} + \frac{2 (- x)}{2 \cdot 4} = \frac{15}{8}$

Rewrite the expression.

$x^{2} + \frac{- x}{4} = \frac{15}{8}$

Move the negative in front of the fraction.

$x^{2} - \frac{x}{4} = \frac{15}{8}$

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 $b$ .

${(\frac{b}{2})}^{2} = {(- \frac{1}{8})}^{2}$

Step 4

Add the term to each side of the equation.

$x^{2} - \frac{x}{4} + {(- \frac{1}{8})}^{2} = \frac{15}{8} + {(- \frac{1}{8})}^{2}$

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 $- \frac{1}{8}$ .

$x^{2} - \frac{x}{4} + {(- 1)}^{2} {(\frac{1}{8})}^{2} = \frac{15}{8} + {(- \frac{1}{8})}^{2}$

Apply the product rule to $\frac{1}{8}$ .

$x^{2} - \frac{x}{4} + {(- 1)}^{2} \frac{1^{2}}{8^{2}} = \frac{15}{8} + {(- \frac{1}{8})}^{2}$

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

$x^{2} - \frac{x}{4} + 1 \frac{1^{2}}{8^{2}} = \frac{15}{8} + {(- \frac{1}{8})}^{2}$

Multiply $\frac{1^{2}}{8^{2}}$ by $1$ .

$x^{2} - \frac{x}{4} + \frac{1^{2}}{8^{2}} = \frac{15}{8} + {(- \frac{1}{8})}^{2}$

One to any power is one.

$x^{2} - \frac{x}{4} + \frac{1}{8^{2}} = \frac{15}{8} + {(- \frac{1}{8})}^{2}$

Raise $8$ to the power of $2$ .

$x^{2} - \frac{x}{4} + \frac{1}{64} = \frac{15}{8} + {(- \frac{1}{8})}^{2}$

Simplify the right side.

Simplify $\frac{15}{8} + {(- \frac{1}{8})}^{2}$ .

Simplify each term.

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

Apply the product rule to $- \frac{1}{8}$ .

$x^{2} - \frac{x}{4} + \frac{1}{64} = \frac{15}{8} + {(- 1)}^{2} {(\frac{1}{8})}^{2}$

Apply the product rule to $\frac{1}{8}$ .

$x^{2} - \frac{x}{4} + \frac{1}{64} = \frac{15}{8} + {(- 1)}^{2} \frac{1^{2}}{8^{2}}$

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

$x^{2} - \frac{x}{4} + \frac{1}{64} = \frac{15}{8} + 1 \frac{1^{2}}{8^{2}}$

Multiply $\frac{1^{2}}{8^{2}}$ by $1$ .

$x^{2} - \frac{x}{4} + \frac{1}{64} = \frac{15}{8} + \frac{1^{2}}{8^{2}}$

One to any power is one.

$x^{2} - \frac{x}{4} + \frac{1}{64} = \frac{15}{8} + \frac{1}{8^{2}}$

Raise $8$ to the power of $2$ .

$x^{2} - \frac{x}{4} + \frac{1}{64} = \frac{15}{8} + \frac{1}{64}$

To write $\frac{15}{8}$ as a fraction with a common denominator, multiply by $\frac{8}{8}$ .

$x^{2} - \frac{x}{4} + \frac{1}{64} = \frac{15}{8} \cdot \frac{8}{8} + \frac{1}{64}$

Write each expression with a common denominator of $64$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{15}{8}$ by $\frac{8}{8}$ .

$x^{2} - \frac{x}{4} + \frac{1}{64} = \frac{15 \cdot 8}{8 \cdot 8} + \frac{1}{64}$

Multiply $8$ by $8$ .

$x^{2} - \frac{x}{4} + \frac{1}{64} = \frac{15 \cdot 8}{64} + \frac{1}{64}$

Combine the numerators over the common denominator.

$x^{2} - \frac{x}{4} + \frac{1}{64} = \frac{15 \cdot 8 + 1}{64}$

Simplify the numerator.

Multiply $15$ by $8$ .

$x^{2} - \frac{x}{4} + \frac{1}{64} = \frac{120 + 1}{64}$

Add $120$ and $1$ .

$x^{2} - \frac{x}{4} + \frac{1}{64} = \frac{121}{64}$

Step 6

Factor the perfect trinomial square into ${(x - \frac{1}{8})}^{2}$ .

${(x - \frac{1}{8})}^{2} = \frac{121}{64}$

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.

$x - \frac{1}{8} = \pm \sqrt{\frac{121}{64}}$

Simplify $\pm \sqrt{\frac{121}{64}}$ .

Rewrite $\sqrt{\frac{121}{64}}$ as $\frac{\sqrt{121}}{\sqrt{64}}$ .

$x - \frac{1}{8} = \pm \frac{\sqrt{121}}{\sqrt{64}}$

Simplify the numerator.

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

$x - \frac{1}{8} = \pm \frac{\sqrt{11^{2}}}{\sqrt{64}}$

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

$x - \frac{1}{8} = \pm \frac{11}{\sqrt{64}}$

Simplify the denominator.

Rewrite $64$ as $8^{2}$ .

$x - \frac{1}{8} = \pm \frac{11}{\sqrt{8^{2}}}$

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

$x - \frac{1}{8} = \pm \frac{11}{8}$

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.

$x - \frac{1}{8} = \frac{11}{8}$

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

Add $\frac{1}{8}$ to both sides of the equation.

$x = \frac{11}{8} + \frac{1}{8}$

Combine the numerators over the common denominator.

$x = \frac{11 + 1}{8}$

Add $11$ and $1$ .

$x = \frac{12}{8}$

Cancel the common factor of $12$ and $8$ .

Factor $4$ out of $12$ .

$x = \frac{4 (3)}{8}$

Cancel the common factors.

Factor $4$ out of $8$ .

$x = \frac{4 \cdot 3}{4 \cdot 2}$

Cancel the common factor.

$x = \frac{4 \cdot 3}{4 \cdot 2}$

Rewrite the expression.

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

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

$x - \frac{1}{8} = - \frac{11}{8}$

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

Add $\frac{1}{8}$ to both sides of the equation.

$x = - \frac{11}{8} + \frac{1}{8}$

Combine the numerators over the common denominator.

$x = \frac{- 11 + 1}{8}$

Add $- 11$ and $1$ .

$x = \frac{- 10}{8}$

Cancel the common factor of $- 10$ and $8$ .

Factor $2$ out of $- 10$ .

$x = \frac{2 (- 5)}{8}$

Cancel the common factors.

Factor $2$ out of $8$ .

$x = \frac{2 \cdot - 5}{2 \cdot 4}$

Cancel the common factor.

$x = \frac{2 \cdot - 5}{2 \cdot 4}$

Rewrite the expression.

$x = \frac{- 5}{4}$

Move the negative in front of the fraction.

$x = - \frac{5}{4}$

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

$x = \frac{3}{2}, - \frac{5}{4}$

Solve by Completing the Square 8x^2-2x-15=0

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