Solve by Completing the Square 5x^2-x-4=0

$5 x^{2} - x - 4 = 0$

Step 1

Add $4$ to both sides of the equation.

$5 x^{2} - x = 4$

Step 2

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

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

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

Simplify the left side.

Simplify each term.

Cancel the common factor of $5$ .

Cancel the common factor.

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

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

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

Move the negative in front of the fraction.

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

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 $b$ .

${(\frac{b}{2})}^{2} = {(- \frac{1}{10})}^{2}$

Step 4

Add the term to each side of the equation.

$x^{2} - \frac{x}{5} + {(- \frac{1}{10})}^{2} = \frac{4}{5} + {(- \frac{1}{10})}^{2}$

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 $- \frac{1}{10}$ .

$x^{2} - \frac{x}{5} + {(- 1)}^{2} {(\frac{1}{10})}^{2} = \frac{4}{5} + {(- \frac{1}{10})}^{2}$

Apply the product rule to $\frac{1}{10}$ .

$x^{2} - \frac{x}{5} + {(- 1)}^{2} \frac{1^{2}}{10^{2}} = \frac{4}{5} + {(- \frac{1}{10})}^{2}$

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

$x^{2} - \frac{x}{5} + 1 \frac{1^{2}}{10^{2}} = \frac{4}{5} + {(- \frac{1}{10})}^{2}$

Multiply $\frac{1^{2}}{10^{2}}$ by $1$ .

$x^{2} - \frac{x}{5} + \frac{1^{2}}{10^{2}} = \frac{4}{5} + {(- \frac{1}{10})}^{2}$

One to any power is one.

$x^{2} - \frac{x}{5} + \frac{1}{10^{2}} = \frac{4}{5} + {(- \frac{1}{10})}^{2}$

Raise $10$ to the power of $2$ .

$x^{2} - \frac{x}{5} + \frac{1}{100} = \frac{4}{5} + {(- \frac{1}{10})}^{2}$

Simplify the right side.

Simplify $\frac{4}{5} + {(- \frac{1}{10})}^{2}$ .

Simplify each term.

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

Apply the product rule to $- \frac{1}{10}$ .

$x^{2} - \frac{x}{5} + \frac{1}{100} = \frac{4}{5} + {(- 1)}^{2} {(\frac{1}{10})}^{2}$

Apply the product rule to $\frac{1}{10}$ .

$x^{2} - \frac{x}{5} + \frac{1}{100} = \frac{4}{5} + {(- 1)}^{2} \frac{1^{2}}{10^{2}}$

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

$x^{2} - \frac{x}{5} + \frac{1}{100} = \frac{4}{5} + 1 \frac{1^{2}}{10^{2}}$

Multiply $\frac{1^{2}}{10^{2}}$ by $1$ .

$x^{2} - \frac{x}{5} + \frac{1}{100} = \frac{4}{5} + \frac{1^{2}}{10^{2}}$

One to any power is one.

$x^{2} - \frac{x}{5} + \frac{1}{100} = \frac{4}{5} + \frac{1}{10^{2}}$

Raise $10$ to the power of $2$ .

$x^{2} - \frac{x}{5} + \frac{1}{100} = \frac{4}{5} + \frac{1}{100}$

To write $\frac{4}{5}$ as a fraction with a common denominator, multiply by $\frac{20}{20}$ .

$x^{2} - \frac{x}{5} + \frac{1}{100} = \frac{4}{5} \cdot \frac{20}{20} + \frac{1}{100}$

Write each expression with a common denominator of $100$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{4}{5}$ by $\frac{20}{20}$ .

$x^{2} - \frac{x}{5} + \frac{1}{100} = \frac{4 \cdot 20}{5 \cdot 20} + \frac{1}{100}$

Multiply $5$ by $20$ .

$x^{2} - \frac{x}{5} + \frac{1}{100} = \frac{4 \cdot 20}{100} + \frac{1}{100}$

Combine the numerators over the common denominator.

$x^{2} - \frac{x}{5} + \frac{1}{100} = \frac{4 \cdot 20 + 1}{100}$

Simplify the numerator.

Multiply $4$ by $20$ .

$x^{2} - \frac{x}{5} + \frac{1}{100} = \frac{80 + 1}{100}$

Add $80$ and $1$ .

$x^{2} - \frac{x}{5} + \frac{1}{100} = \frac{81}{100}$

Step 6

Factor the perfect trinomial square into ${(x - \frac{1}{10})}^{2}$ .

${(x - \frac{1}{10})}^{2} = \frac{81}{100}$

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.

$x - \frac{1}{10} = \pm \sqrt{\frac{81}{100}}$

Simplify $\pm \sqrt{\frac{81}{100}}$ .

Rewrite $\sqrt{\frac{81}{100}}$ as $\frac{\sqrt{81}}{\sqrt{100}}$ .

$x - \frac{1}{10} = \pm \frac{\sqrt{81}}{\sqrt{100}}$

Simplify the numerator.

Rewrite $81$ as $9^{2}$ .

$x - \frac{1}{10} = \pm \frac{\sqrt{9^{2}}}{\sqrt{100}}$

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

$x - \frac{1}{10} = \pm \frac{9}{\sqrt{100}}$

Simplify the denominator.

Rewrite $100$ as $10^{2}$ .

$x - \frac{1}{10} = \pm \frac{9}{\sqrt{10^{2}}}$

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

$x - \frac{1}{10} = \pm \frac{9}{10}$

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.

$x - \frac{1}{10} = \frac{9}{10}$

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

Add $\frac{1}{10}$ to both sides of the equation.

$x = \frac{9}{10} + \frac{1}{10}$

Combine the numerators over the common denominator.

$x = \frac{9 + 1}{10}$

Add $9$ and $1$ .

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

Divide $10$ by $10$ .

$x = 1$

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

$x - \frac{1}{10} = - \frac{9}{10}$

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

Add $\frac{1}{10}$ to both sides of the equation.

$x = - \frac{9}{10} + \frac{1}{10}$

Combine the numerators over the common denominator.

$x = \frac{- 9 + 1}{10}$

Add $- 9$ and $1$ .

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

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

Factor $2$ out of $- 8$ .

$x = \frac{2 (- 4)}{10}$

Cancel the common factors.

Factor $2$ out of $10$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

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

$x = 1, - \frac{4}{5}$

Solve by Completing the Square 5x^2-x-4=0

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