Solve by Completing the Square 8x^2+4x-4=0

$8 x^{2} + 4 x - 4 = 0$

Step 1

Add $4$ to both sides of the equation.

$8 x^{2} + 4 x = 4$

Step 2

Divide each term in $8 x^{2} + 4 x = 4$ by $8$ and simplify.

Divide each term in $8 x^{2} + 4 x = 4$ by $8$ .

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

Simplify the left side.

Simplify each term.

Cancel the common factor of $8$ .

Cancel the common factor.

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

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

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

Cancel the common factor of $4$ and $8$ .

Factor $4$ out of $4 x$ .

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

Cancel the common factors.

Factor $4$ out of $8$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the right side.

Cancel the common factor of $4$ and $8$ .

Factor $4$ out of $4$ .

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

Cancel the common factors.

Factor $4$ out of $8$ .

$x^{2} + \frac{x}{2} = \frac{4 \cdot 1}{4 \cdot 2}$

Cancel the common factor.

$x^{2} + \frac{x}{2} = \frac{4 \cdot 1}{4 \cdot 2}$

Rewrite the expression.

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

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 $b$ .

${(\frac{b}{2})}^{2} = {(\frac{1}{4})}^{2}$

Step 4

Add the term to each side of the equation.

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

Step 5

Simplify the equation.

Simplify the left side.

Simplify each term.

Apply the product rule to $\frac{1}{4}$ .

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

One to any power is one.

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

Raise $4$ to the power of $2$ .

$x^{2} + \frac{x}{2} + \frac{1}{16} = \frac{1}{2} + {(\frac{1}{4})}^{2}$

Simplify the right side.

Simplify $\frac{1}{2} + {(\frac{1}{4})}^{2}$ .

Simplify each term.

Apply the product rule to $\frac{1}{4}$ .

$x^{2} + \frac{x}{2} + \frac{1}{16} = \frac{1}{2} + \frac{1^{2}}{4^{2}}$

One to any power is one.

$x^{2} + \frac{x}{2} + \frac{1}{16} = \frac{1}{2} + \frac{1}{4^{2}}$

Raise $4$ to the power of $2$ .

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

To write $\frac{1}{2}$ as a fraction with a common denominator, multiply by $\frac{8}{8}$ .

$x^{2} + \frac{x}{2} + \frac{1}{16} = \frac{1}{2} \cdot \frac{8}{8} + \frac{1}{16}$

Write each expression with a common denominator of $16$ , by multiplying each by an appropriate factor of $1$ .

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

$x^{2} + \frac{x}{2} + \frac{1}{16} = \frac{8}{2 \cdot 8} + \frac{1}{16}$

Multiply $2$ by $8$ .

$x^{2} + \frac{x}{2} + \frac{1}{16} = \frac{8}{16} + \frac{1}{16}$

Combine the numerators over the common denominator.

$x^{2} + \frac{x}{2} + \frac{1}{16} = \frac{8 + 1}{16}$

Add $8$ and $1$ .

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

Step 6

Factor the perfect trinomial square into ${(x + \frac{1}{4})}^{2}$ .

${(x + \frac{1}{4})}^{2} = \frac{9}{16}$

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.

$x + \frac{1}{4} = \pm \sqrt{\frac{9}{16}}$

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

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

$x + \frac{1}{4} = \pm \frac{\sqrt{9}}{\sqrt{16}}$

Simplify the numerator.

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

$x + \frac{1}{4} = \pm \frac{\sqrt{3^{2}}}{\sqrt{16}}$

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

$x + \frac{1}{4} = \pm \frac{3}{\sqrt{16}}$

Simplify the denominator.

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

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

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

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

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.

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

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

Subtract $\frac{1}{4}$ from both sides of the equation.

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

Combine the numerators over the common denominator.

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

Subtract $1$ from $3$ .

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

Cancel the common factor of $2$ and $4$ .

Factor $2$ out of $2$ .

$x = \frac{2 (1)}{4}$

Cancel the common factors.

Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

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

Subtract $\frac{1}{4}$ from both sides of the equation.

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

Combine the numerators over the common denominator.

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

Subtract $1$ from $- 3$ .

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

Divide $- 4$ by $4$ .

$x = - 1$

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

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

Solve by Completing the Square 8x^2+4x-4=0

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