We want to solve the quadratic equation by completing the square. To do so we will start by rewriting the equation so all terms with x are on one side of the equation and all constants are on the other side.
x^2+14x+40=- 5 ⇔ x^2+14x=- 45
In a quadratic expression b is the linear coefficient. For the equation above, we have that b=14. Let's now calculate ( b2 )^2.
The solutions for this equation are x=- 7± 2. Let's separate them into the positive and negative cases.
x=- 7± 2
x_1=- 7+ 2
x_2=- 7- 2
x_1=- 5
x_2=- 9
By completing the square we found that the solutions are x_1=- 5 and x_2=- 9.
Using Zero Product Property
Now, we want to solve the given equation using the Zero Product Property. To do this we need to rewrite the given equation in the factored form.
Let's start by gathering all terms on the left-hand side of the equation
x^2+14x+40=- 5
⇕
x^2+14x+45=0
Now, with all terms on one side of the equation, we can continue by factoring.
Factoring
To factor a trinomial with a leading coefficient of 1, think of the process as multiplying two binomials in reverse. Let's start by taking a look at the constant term.
x^2+14x+45=0
In this case, we have 45. This is a positive number, so for the product of the constant terms in the factors to be positive, these constants must have the same sign (both positive or both negative.)
Factor Constants
Product of Constants
1 and 45
45
-1 and -45
45
3 and 15
45
-3 and -15
45
5 and 9
45
-5 and -9
45
Next, let's consider the coefficient of the linear term.
x^2+14n+45=0
For this term we need the sum of the factors that produced the constant term to equal the coefficient of the linear term, 14.
Factors
Sum of Factors
1 and 45
46
-1 and -45
-46
3 and 15
18
-3 and -15
-18
5 and 9
14
-5 and -9
-14
We found the factors whose product is 45 and whose sum is 14.
x^2+14x+45=0
⇕
(x+5)(x+9)=0
Now the equation is written in a factored form.
Zero Product Property
Since the equation is already written in factored form, we can now use the Zero Product Property.
We will use the Quadratic Formula to solve the given quadratic equation.
ax^2+ bx+ c=0 ⇕ x=- b± sqrt(b^2-4 a c)/2 a
We first need to identify the values of a, b, and c. Let's recall the rewritten form of the given equation, obtained in Part B.
1x^2+ 14x+ 45=0
We see that a= 2, b= 14, and c= 45, so we will substitute these values into the Quadratic Formula.
The solutions for this equation are x=- 7± 2. Let's separate them into the positive and negative cases.
x=- 7± 2
x_1=- 7+ 2
x_2=- 7- 2
x_1=- 5
x_2=- 9
Using the Quadratic Formula, we found that the solutions are x_1=- 5 and x_2=- 9. Notice that those are the same solutions as obtained previously by completing the square, and then by factoring and using Zero Product Property. All three methods are equally good, so they give the same solutions.
