We can solve an equation of the form

x^{2} + b x = c,

where

b

and

c

are real numbers, by completing the square. Notice that we can always rewrite any quadratic equation in this form by isolating the variable terms and dividing by the coefficient of

x^{2}

if needed. Let's see an example.

4 x^{2} + 24 x + 32 = 0

SubEqn

$LHS - 32 = RHS - 32$

4 x^{2} + 24 x = - 32

DivEqn

$LHS / 4 = RHS / 4$

x^{2} + 6 x = - 8

Now we need to ensure that we have an expression that is a perfect square trinomial. Then, we will be able to factor it as the square of a binomial and take the square root on both sides to find the solutions. Let's recall the expansion of the square of a binomial.

(x + a)^{2} = x^{2} + 2 a x + a^{2}

By adding the appropriate constant, we can obtain a perfect square trinomial. Let's compare the right-hand side of this formula to the left-hand side of our expression.

x^{2} + 2 a x + a^{2} x^{2} + b x + ?

Notice that the

b

parameter of our expression corresponds to the factor

2 a

in the expansion for the square of the binomial. We can find the quantity corresponding to

a^{2}

that we would have to add from this relationship by solving for

a

first and then squaring both sides.

2 a = b

DivEqn

$LHS / 2 = RHS / 2$

a = \frac{b}{2}

RaiseEqn

$LHS^{2} = RHS^{2}$

a^{2} = (\frac{b}{2})^{2}

This means that by adding the term

(\frac{b}{2})^{2}

we can get a perfect square trinomial! Let's try this with the example equation we had at the beginning,

x^{2} + 6 x = - 8 .

Notice that here

b = 6 .

b = 6

DivEqn

$LHS / 2 = RHS / 2$

\frac{b}{2} = 3

RaiseEqn

$LHS^{2} = RHS^{2}$

(\frac{b}{2})^{2} = 9

Then, adding

9

to both sides of

x^{2} + 6 x = - 8

will let us rewrite the left-hand side as the square of a binomial.

x^{2} + 6 x = - 8

AddEqn

$LHS + 9 = RHS + 9$

x^{2} + 6 x + 9 = 1

FacPosPerfectSquare

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

(x + 3)^{2} = 1

SqrtEqn

$LHS = RHS$

x + 3 = \pm 1

SubEqn

$LHS - 3 = RHS - 3$

x = - 3 \pm 1

As we can see, the solutions for this equation are

x_{1} = - 3 + 1 = - 2

and

x_{2} = - 3 - 1 = - 4 .

Summary of the Procedure

As we saw above, we can complete the square by following the next steps.

We isolate the terms containing the variable $x .$
We divide both sides by the coefficient of $x^{2} .$
We identify the coefficient of the linear term as $b .$
We add $(\frac{b}{2})^{2}$ to both sides.

Exercise

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