Exercise 58 Page 588

We will compare and contrast the four strategies for solving quadratic equations: completing the square, graphing, factoring, and using the discriminant. It will be easier to visualize these methods if we choose a quadratic equation to model them with, so let's use the following.

x^{2} - 5 x - 7 = 0

Completing the Square

Completing the square is a technique for converting a quadratic polynomial of the form

x^{2} + b x

to the form

(x - h)^{2} + k

for some values of

h

and

k .

It can be done by adding

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

to

x^{2} + b x .

Consider the given equation.

x^{2} - 5 x - 7 = 0

First, let's add

7

to both sides to get all the terms with

x

on the left-hand side and the other terms on the right-hand side of the equation.

x^{2} - 5 x - 7 = 0 \Leftrightarrow x^{2} - 5 x = 7

For this equation, we have that

b = - 5 .

Let's now calculate

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

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

▼

Substitute

- 5

for

b

and evaluate

Substitute

$b = - 5$

(\frac{- 5}{2})^{2}

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

\frac{( - 5 ) ^{2}}{2 ^{2}}

NegBaseToPosBase

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

\frac{5 ^{2}}{2 ^{2}}

CalcPow

Calculate power

\frac{2 5}{4}

Next, we will add

(\frac{b}{2})^{2} = \frac{2 5}{4}

to both sides of our equation. Then we will factor the trinomial on the left-hand side using the formula for perfect square trinomials and solve the equation.

x^{2} - 5 x = 7

AddEqn

$LHS + \frac{2 5}{4} = RHS + \frac{2 5}{4}$

x^{2} - 5 x + \frac{2 5}{4} = 7 + \frac{2 5}{4}

Rewrite

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

x^{2} - (2 \cdot \frac{5}{2}) x + \frac{2 5}{4} = 7 + \frac{2 5}{4}

Rewrite

Rewrite $\frac{2 5}{4}$ as $(\frac{5}{2})^{2}$

x^{2} - (2 \cdot \frac{5}{2}) x + (\frac{5}{2})^{2} = 7 + \frac{2 5}{4}

FacNegPerfectSquare

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

(x - \frac{5}{2})^{2} = 7 + \frac{2 5}{4}

▼

Solve for

x

NumberToFrac

$a = \frac{4 \cdot a}{4}$

(x - \frac{5}{2})^{2} = \frac{2 8}{4} + \frac{2 5}{4}

AddFrac

Add fractions

(x - \frac{5}{2})^{2} = \frac{5 3}{4}

SqrtEqn

$LHS = RHS$

(x - \frac{5}{2})^{2} = \frac{5 3}{4}

CalcRoot

Calculate root

x - \frac{5}{2} = \pm \frac{5 3}{4}

SqrtQuot

$\frac{a}{b} = \frac{a}{b}$

x - \frac{5}{2} = \pm \frac{5 3}{2}

AddEqn

$LHS + \frac{5}{2} = RHS + \frac{5}{2}$

x = \frac{5}{2} \pm \frac{5 3}{2}

Both

x = \frac{5}{2} + \frac{5 3}{2}

and

x = \frac{5}{2} - \frac{5 3}{2}

are solutions of the equation. Note that this method always give us the exact solutions.

Graphing

Now let's try graphing. In this method the first step is to plot the quadratic function,

y = x^{2} - 5 x - 7,

on a coordinate plane.

The roots of a quadratic equation are the zeros of a quadratic function. Therefore, we should approximate where the graph crosses the

x -

axis.

Therefore, the roots of the equation $x^{2} - 5 x - 7 = 0$ are approximately $- 1.1$ and $6.1 .$ Note that by this method we are not able to find the exact solutions and usually we will spend a lot of time plotting the parabola in a coordinate system.

Factoring

To solve the given equation by factoring, we will start by identifying the values of

a,

b,

and

c .

x^{2} - 5 - 7 x = 0 ⇕ 1 x^{2} + (- 5) x + (- 7) = 0

We have a quadratic equation with

a = 1,

b = - 5,

and

c = - 7 .

To factor the left-hand side we need to find a factor pair of

1 \times (- 7) = - 7

whose sum is

- 5 .

Since

- 7

is a negative number, we will only consider factors with opposite signs — one positive and one negative — so that their product is negative.

Factor Pair	Product of Factors	Sum of Factors
$1$ and $- 7$	$- 7^{1 \times (- 7)}$	$- 6 1 + (- 7)$
$- 1$ and $7$	$- 7^{- 1 \times 7}$	$6 - 1 + 7$

Unfortunately, we did not find the integers whose product is $- 7$ and whose sum is $- 5 .$ Therefore, we cannot continue with this strategy. Factoring does not always work — especially when the roots are not rational numbers, like in our case. Note that if the coefficient $c$ was a big number we must do a lot of calculations.

Discriminant

We will use the Quadratic Formula to solve the given quadratic equation.

a x^{2} + b x + c = 0 ⇕ x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

We first need to identify the values of

a,

b,

and

c .

x^{2} - 5 x - 7 = 0 ⇕ 1 x^{2} + (- 5) x + (- 7) = 0

We see that

a = 1,

b = - 5,

and

c = - 7 .

Let's substitute these values into the Quadratic Formula.

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

SubstituteValues

Substitute values

x = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( 1 ) ( - 7 )}{2 ( 1 )}

▼

Simplify

NegNeg

$- (- a) = a$

x = \frac{5 \pm ( - 5 ) ^{2} - 4 ( 1 ) ( - 7 )}{2 ( 1 )}

CalcPow

Calculate power

x = \frac{5 \pm 2 5 - 4 ( 1 ) ( - 7 )}{2 ( 1 )}

Multiply

x = \frac{5 \pm 2 5 - 4 ( - 7 )}{2}

MultNegNegOnePar

$- a (- b) = a \cdot b$

x = \frac{5 \pm 2 5 + 2 8}{2}

AddTerms

Add terms

x = \frac{5 \pm 5 3}{2}

The solutions for this equation are

x = \frac{5 \pm 5 3}{2} .

Let's separate them into the positive and negative cases.

$x = \frac{5 \pm 5 3}{2}$
$x_{1} = \frac{5 + 5 3}{2}$	$x_{2} = \frac{5 - 5 3}{2}$

Using the Quadratic Formula, we found that the solutions of the given equation are $x_{1} = \frac{5 + 5 3}{2}$ and $x_{2} = \frac{5 - 5 3}{2} .$ Note that this method always gives us the exact solutions.

Conclusion

Now we will describe the advantages and disadvantages of each method. Unfortunately, not all the methods are equivalent.

Method	Type	Exact	Fast
Completing the Square	Algebraic	$Yes$	$No$
Graphing	Geometric	$No$	$No$
Factoring	Algebraic	$No$	$No$
Distriminant	Algebraic	$Yes$	$Yes$

In general, using discriminant is the best. It is usually the fastest and always exact.