Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
2. Section 3.2
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Exercise 117 Page 158

Write 14x as 11x+3x to factor the left-hand side. Add 7^2 on both sides to complete the square.

x=-3 and x=-11

Practice makes perfect

Let's solve the equation twice! One time we will use the Zero Product Property and the other time we will complete the square.

Zero Product Property

To solve this equation with the Zero Product Property, we must first factor the left-hand side. To do that, we can use a generic rectangle and a diamond problem. We know that x^2 and 33 on the left-hand side goes into the lower left and upper right corner of the generic rectangle.

To fill in the remaining two corners, we need to find two x-terms that sum to 14x and have a product of 33x^2.

Notice that the product and sum are both positive. This means both factors must be positive. |c|c|c|c|c| [-1em] Product & ax(bx) & ax+bx & Sum & 14x? [0.2em] [-1em] 33x^2 & 3x(11x) & 3x+11x & 14x & âś“ [0.3em] When one factor is 3x and the other is 11x we have a product of 33x^2 and a sum of 14x. Now we can complete the diamond and generic rectangle.

To factor the left-hand side, we add each side of the generic rectangle and multiply the sums. x^2+14x+33=0 ⇕ (x+3)(x+11)=0 Now we can solve the equation with the Zero Product Property.
(x+3)(x+11)=0
lcx+3=0 & (I) x+11=0 & (II)
lx=-3 x+11=0
lx_1=-3 x_2=-11
x=-3 and x=-11 solves the equation.

Completing the Square

Let's now try a different method. When completing the square we look at the x-term, 14x. The coefficient is 14, which means that we need to add ( 142)^2 to both sides of the equation to complete the square.
x^2+14x+33=0
x^2+14x+ (14/2)^2+33= (14/2)^2
â–Ľ
Solve for x
x^2+14x+7^2+33=7^2
x^2+14x+7^2=7^2-33
x^2+2(x)(7)+7^2=7^2-33
(x+7)^2=7^2-33
(x+7)^2=49-33
(x+7)^2=16
x+7=±4
x=-7±4
lcx=-7-4 & (I) x=-7+4 & (II)

(I), (II): Add and subtract terms

lx_1=-11 x_2=-3
We get the solutions x=-3 and x=-11, same as when using the Zero Product Property.