Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
Concept Byte: Quadratic Inequalities
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Exercise 5 Page 257

Practice makes perfect
a To solve the given quadratic inequality, we can start by taking all the terms to the left-hand side and try factoring the expression.
x^2<36
x^2-36<0
x^2-6^2<0
(x+6)(x-6)<0

Since the product of the binomials obtained should be negative, as stated by the inequality symbol (x+6)(x-6) <0, the quantities represented by the binomials should have different signs. Then, we have two options.

Option 1 Option 2
Factors' Signs x+6<0 and x-6>0 x+6>0 and x-6<0
Simplified Form x< - 6 and x>6 x>- 6 and x<6
Solution Set no solution - 6

The solution set for the inequality is - 6points at -6 and 6, as the inequality symbols do not include the boundary values. Then, we shade the values between them.

b We can see that on the left-hand side of the inequality we have the difference of squares. We can start by factoring it to see what the inequality implies for these factors.
x^2 -9 > 0
(x+3)(x-3) > 0

Since the product of the binomials obtained should be positive, as stated by the inequality symbol (x+3)(x-3) >0, the quantities represented by the binomials should have the same signs. We have the two options.

Option 1 Option 2
Factors' Signs x+3>0 and x-3>0 x+3<0 and x-3<0
Simplified Form x> - 3 and x>3 x< - 3 and x<3
Solution Set x>3 x<- 3

The solution set for the inequality is then x< - 3 or x>3. To graph the solution set we need to use two open points at -3 and 3, as the inequality symbols do not include the boundary values. Then, we shade to the left of -3 and to the right of 3.

c Notice that the inequality is asking for numbers whose square is less than - 4. x^2 < - 4 As any negative number becomes positive when squared and positive number remain positive, there are no numbers satisfying this inequality.
d We can start by factoring the expression on the left-hand side to see the implications of the inequality for the factors.
x^2 - 3x - 18 >0 To factor the expression on the left-hand side, we need to find two numbers such that their product is- 18 and their sum is- 3. Since - 18<0, both factors should have different sign. Also, since - 3 < 0, the greater factor should be negative. Let's check for these requirements for different factors of 8.
Factors of - 18 Sum
1,- 18 1-8 = - 7 *
2,- 9 2 - 9 = -7 *
3,- 6 3-6 = - 3 ✓

With this in mind, we can factor our expression. x^2-3x-18 >0 ⇕ (x + 3)(x - 6) >0 Since the product of the binomials obtained should be positive, as stated by the inequality symbol (x+3)(x-6) >0, the quantities represented by the binomials should have the same signs. Then, we have two two options.

Option 1 Option 2
Factors' Signs x+3>0 and x-6>0 x+3<0 and x-6<0
Simplified Form x> - 3 and x>6 x< - 3 and x<6
Solution Set x>6 x<- 3

The solution set for the inequality is then x< - 3 or x>6. To graph the solution set we need to use two open points at -3 and 6, as the inequality symbols do not include the boundary values. Then, we shade to the left of -3 and to the right of 6.