We are asked to use the method of completing the square to solve this equation. This method involves creating a complete square on the left hand side of the equation by adding the square of

half

the

coefficient

of the linear term to each side.

x^{2} + b x = d ⇕ x^{2} + b x + (\frac{b}{2})^{2} = d + (\frac{b}{2})^{2} ⇕ (x + \frac{b}{2})^{2} = d + (\frac{b}{2})^{2}

In our case

b = - 1,

so we add

(\frac{- 1}{2})^{2} = \frac{1}{4}

to each side.

x^{2} - x = 1

AddEqn

$LHS + \frac{1}{4} = RHS + \frac{1}{4}$

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

FacNegPerfectSquare

$a^{2} - 2 a b + b^{2} = (a - b)^{2}$

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

AddTerms

Add terms

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

We can now use the Square Root Property to finish solving the equation.

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

SqrtEqn

$LHS = RHS$

x - \frac{1}{2} = \pm \frac{5}{4}

AddEqn

$LHS + \frac{1}{2} = RHS + \frac{1}{2}$

x = \frac{1}{2} \pm \frac{5}{4}

▼

Simplify right-hand side

Rewrite

Rewrite $\frac{5}{4}$ as $\frac{5}{2}$

x = \frac{1}{2} \pm \frac{5}{2}

AddFrac

Add fractions

x = \frac{1 \pm 5}{2}

Since

x

represents a length, it is certainly not negative.

x = \frac{1 + 5}{2}

Exercise

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