Solve by Completing the Square x^2-x=7/4

$x^{2} - x = \frac{7}{4}$

Step 1

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

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

Step 2

Add the term to each side of the equation.

$x^{2} - x + {(- \frac{1}{2})}^{2} = \frac{7}{4} + {(- \frac{1}{2})}^{2}$

Step 3

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 $- \frac{1}{2}$ .

$x^{2} - x + {(- 1)}^{2} {(\frac{1}{2})}^{2} = \frac{7}{4} + {(- \frac{1}{2})}^{2}$

Apply the product rule to $\frac{1}{2}$ .

$x^{2} - x + {(- 1)}^{2} \frac{1^{2}}{2^{2}} = \frac{7}{4} + {(- \frac{1}{2})}^{2}$

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

$x^{2} - x + 1 \frac{1^{2}}{2^{2}} = \frac{7}{4} + {(- \frac{1}{2})}^{2}$

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

$x^{2} - x + \frac{1^{2}}{2^{2}} = \frac{7}{4} + {(- \frac{1}{2})}^{2}$

One to any power is one.

$x^{2} - x + \frac{1}{2^{2}} = \frac{7}{4} + {(- \frac{1}{2})}^{2}$

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

$x^{2} - x + \frac{1}{4} = \frac{7}{4} + {(- \frac{1}{2})}^{2}$

Simplify the right side.

Simplify $\frac{7}{4} + {(- \frac{1}{2})}^{2}$ .

Simplify each term.

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

Apply the product rule to $- \frac{1}{2}$ .

$x^{2} - x + \frac{1}{4} = \frac{7}{4} + {(- 1)}^{2} {(\frac{1}{2})}^{2}$

Apply the product rule to $\frac{1}{2}$ .

$x^{2} - x + \frac{1}{4} = \frac{7}{4} + {(- 1)}^{2} \frac{1^{2}}{2^{2}}$

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

$x^{2} - x + \frac{1}{4} = \frac{7}{4} + 1 \frac{1^{2}}{2^{2}}$

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

$x^{2} - x + \frac{1}{4} = \frac{7}{4} + \frac{1^{2}}{2^{2}}$

One to any power is one.

$x^{2} - x + \frac{1}{4} = \frac{7}{4} + \frac{1}{2^{2}}$

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

$x^{2} - x + \frac{1}{4} = \frac{7}{4} + \frac{1}{4}$

Combine fractions.

Combine the numerators over the common denominator.

$x^{2} - x + \frac{1}{4} = \frac{7 + 1}{4}$

Simplify the expression.

Add $7$ and $1$ .

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

Divide $8$ by $4$ .

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

Step 4

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

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

Step 5

Solve the equation for $x$ .

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

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

Add $\frac{1}{2}$ to both sides of the equation.

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

Step 6

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = 1.91421356 \dots, - 0.91421356 \dots$

Solve by Completing the Square x^2-x=7/4

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