Core Connections Algebra 1, 2013
CC
Core Connections Algebra 1, 2013 View details
3. Section 11.3
Continue to next subchapter

Exercise 83 Page 553

Practice makes perfect
a To solve the given quadratic equation, we will first rewrite the equation so all of the terms are on the left-hand side and then simplify as much as possible. Let's begin by using distribution.
- (x-3)(x+1)=4
(x-3)(x+1)=- 4
x(x-3)+1(x-3)=- 4
x^2-3x+1(x-3)=- 4
â–Ľ
Simplify
x^2-3x+x-3=- 4
x^2-2x-3=- 4
x^2-2x+1=0
Now, look closely at the equation above. It would be much easier to solve if the expression on the left-hand side of the equation was a perfect square trinomial. To determine if an expression is a perfect square trinomial, we need to ask ourselves three questions.
Is the first term a perfect square? x^2= x^2 âś“
Is the last term a perfect square? 1= 1^2 âś“
Is the middle term twice the product of 1 and x? 2x=2* 1* x âś“
As we can see, the answer to all three questions is yes! Therefore, we can write the trinomial as the square of a binomial. Note there is a subtraction sign in the middle. x^2-2x+1=0 ⇔ ( x- 1)^2=0 To solve the above quadratic equation, we will take the square root of both sides of the equation. Since this method gives two solutions — a negative and a positive — remember to consider them both by adding ± to the solution.
(x-1)^2=0
sqrt((x-1)^2)=sqrt(0)
â–Ľ
Solve for x
x-1=± 0
x=1 ± 0
x=1
The solution of this equation is x=1. We can substitute it back into the given equation and simplify to check if our answer is correct.
- (x-3)(x+1)=4
- ( 1-3)( 1+1)? =4
â–Ľ
Simplify
-(- 2)(2)? =4
-(- 4)? =4
4=4 âś“
Substituting and simplifying created a true statement, so we know that x=1 is indeed a solution of the given equation.
b To solve the given inequality, let's first isolate the absolute value on one side. We will do this by dividing both sides of the inequality by 10.
10|x-3| > 40
|x-3|>4
To solve the above inequality, we will create a compound inequality by removing the absolute value. In this case the solution set contains the numbers that make the distance between x and 3 greater than 4 in the positive direction or in the negative direction.

x-3 > 4 or x-3< - 4 Let's isolate x in both of these cases before writing the solution set.

Case 1

x-3 > 4
x> 7
This inequality tells us that all values greater than 7 will satisfy the inequality.

Case 2

x-3 < - 4
x< - 1
This inequality tells us that all values less than - 1 will satisfy the inequality.

Solution Set

The solution to this type of compound inequality is the combination of the solution sets. First Solution Set:& x>7 Second Solution Set:& x<- 1 Combined Solution Set:& x<- 1 or x>7 This inequality tells us that all values greater than 7 or less than - 1 will satisfy the inequality. Now, we can check our solution by substituting four values of x into the original inequality.

Value of x Substitute Simplify
- 2 10| - 2-3|? > 40 50 > 40
0 10| 0-3|? > 40 30 ≯ 40
6 10| 6-3| ? > 40 30 ≯ 40
8 10| 8-3| ? > 40 50 > 40

As we can see, values of x greater than 7 or less than - 1 are indeed the solution of the given inequality.

c We want to solve the given cubic equation. To do this, we will take the cube root of both sides of the given equation.
(x-5)^3=8
(x-5)^3=2^3
sqrt((x-5)^3)=sqrt(2^3)
x-5=2
x=7
We found that the solution for this equation is x=7. We can substitute it back into the given equation and simplify to check if our answer is correct.
(x-5)^3=8
( 7-5)^3? =8
â–Ľ
Simplify
(2)^3? =8
8=8 âś“
Substituting and simplifying created a true statement, so we know that x=7 is indeed a solution of the given equation.