# Solutions for Lesson 12 Practice Questions

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Contents

## Number Lines

For each inequality, find the critical numbers and fill them in at the proper spots on the number line (note that the scale on the number line does not matter, just the relative position of the critical numbers.  Then write in which bracket belongs at each critical position, and check off which regions should be shaded.

a) x^2-25>0

b) 2x^2+5x-12\le 0

## Interval Notation

Solve each inequality for y .  Write solutions in interval notation.

a) 2y^2+6y\le -4

2y^2+6y\le -4
2y^2+6y+4\le 0
y^2+3y+2\le 0
(y+2)(y+1)\le 0
Critical values: y=-2 and y=-1
Test intervals: (-\infty, -2) , (-2, -1) , (-1, \infty)
For (-\infty, -2] , I'll try the value of -5 for y .
2(-5)^2+6(-5)=2(25)-30=50-30=20
20 is not less than or equal to -4 , so the interval (-\infty, -2) is not part of the solution set for y .
For [-2,-1] , I'll try the value of -1.5 for y .
2(-1.5)^2+6(-1.5)=2(2.25)-9=4.5-9=-4.5
-4.5 is less than -4 , satisfying our original inequality, so the interval (-2, -1) is in the solution set for y .
For (-1, \infty) , I'll try the value of 0 for y .
2(0)^2+6(0)=0
0 is not less than or equal to -4 , so the interval (-1, \infty) is not part of the solution set for y .
Now let's test the critical values.
2(-2)^2+6(-2)=2(4)-12=8-12=-4
-4 is equal to -4 , so -1 is part of the solution set.
2(-1)^2+6(-1)=2(1)-6=-4
-4 is equal to -4 , so -1 is part of the solution set.
That means that the solution is the interval [-2, -1]

b) \frac {3y+1}{y-5} > 2

\frac {3y+1}{y-5} > 2
\frac {3y+1}{y-5} -2 > 0
\frac {3y+1}{y-5} -2 \cdot \frac {y-5}{y-5} > 0
\frac {3y+1-2(y-5)}{y-5}>0
\frac {y+11}{y-5}>0
The critical values of y are -11 and 5 .
Test intervals: (-\infty, -11) , (-11, 5) , (5, \infty)
For (-\infty, -11) , I'll try the value of -20 for y .
\frac {3(-20)+1}{(-20)-5}=\frac {-59}{-25}=\frac {59}{25}=2.36
2.36 is greater than 2 , so the interval (-\infty, -11) is part of the solution set for y .
For (-11, 5) , I'll try the value of 0 for y .
\frac {3(0)+1}{(0)-5}=\frac {1}{-5} = -0.2
-0.2 is not greater than 2 , so (-11, 5) is not part of the solution set for y .
For (5, \infty) , I'll try the value of 10 for y .
\frac {3(10)+1}{(10)-5}=\frac {31}{5}=\frac {59}{25}=6.2
6.2 is greater than 2 , so (5, \infty) is part of the solution set for y .
Now let's test the critical values.
\frac {3(-11)+1}{-11-5}=\frac {-32}{-16}=2
2 is not greater than 2 , so -11 is not part of the solution set.
\frac {3(5)+1}{5-5}=\frac {16}{0}
Since the denominator of the fraction is 0 , and we can't divide by 0 , 5 is not part of the solution set.
That means that the solution is (-\infty,-11)\union (5,\infty)

Solve each inequality for x , and write your answers in inequality form.
a) x^2\ge 9

x^2\ge 9
x^2-9\ge 0
(x-3)(x+3)\ge 0
Critical values: -3 and 3
Test intervals: x< -3 , -3< x< 3 , x> 3
Testing x=-5 for x< -3 :
(-5)^2=25
25 is greater than 9 , so x< -3 is part of the solution set.
Testing x=0 for -3< x< 3 : (0)^2=0
0 is not greater than or equal to 9 , so -3< x< 3 is not part of the solution set.
Testing x=5 for x>3 :
(5)^2=25
25 is greater than 9 , so x>3 is part of the solution set.
Now let's test the critical values, -3 and 3 .
(-3)^2=9
9 is equal to 9 , so -3 is part of the solution set.
(3)^2=9
9 is equal to 9 , so 3 is part of the solution set.
That means that the solution is -3\le x\le 3

b) x^2+5x-2\le -8

x^2+5x-2\le -8
x^2+5x+6\le 0
(x+3)(x+2)\le 0
Critical values: x=-3 and x=-2
Test intervals: x<-3 , -3<x <-2 , -2<x
Testing x=-5 for x<-3 :
(-5)^2+5(-5)+6=25-25+6=6
6 is not less than or equal to 0 , so -3>x is not part of the solution set.
Testing -2.5 for -3 < x<-2 :
(-2.5)^2+5(-2.5)+6=6.25-12.5+6=-0.25
-0.25 is less than 0 , so -3<x<-2 is part of the solution set.
Testing 0 for -2<x :
(0)^2+5(0)+6

c) 2x^2+4x+1>-1

2x^2+4x+1>-1
2x^2+4x+2>0
x^2+2x+1>0
(x+1)^2>0
Critical values: x=-1
Test intervals: x<-1 , x>-1
Testing -2 for x<-1 :
2(-2)+4(-2)+1=-4-8+1=-3
-3 is not greater than -1 , so x<-1 is not part of the solution set.
Testing 0 for x>-1 :
2(0)^2+4(0)+1=0+0+1=1
1 is greater than -1 , so x>-1 is part of the solution set.
Testing the critical value, x=-1 :
2(-1)^2+4(-1)+1=2(1)-4+1=-1
-1 is not greater than -1 , so -1 is not part of the solution set.
That means that the solution is x>-1

d) -x^2+3x+4<0

-x^2+3x+4<0
x^2-3x-4>0
(x-4)(x+1)>0
Critical values: x=4 and x=-1
Test intervals: x<-1 , -1<x<4 , x>4
Testing x=-2 for x<-1 :
-(-2)^2+3(-2)+4=-4-6+4=-6
-6 is less than 0 , so x<-1 is part of the solution set.
Testing x=0 for -1<x<4 :
-(0)^2+3(0)+4=4
4 is not less than 0 , so -1<x<4 is not part of the solution set.
Testing x=5 for x>4 :
-(5)^2+3(5)+4=-25+15+4=-6
-6 is less than 0 , so x>4 is part of the solution set.
Testing the critical values, x=-1 and x=4 :
-(4)^2+3(4)+4=-16+12+4=0
-(-1)^2+3(-1)+4=-1-3+4=0
0 is not less than 0 , so neither 4 nor -1 is part of the solution set.
That means that the solution is x<-1\or x>4

e) 3x^2< -21x+24

3x^2< -21x+24
3x^2+21x-24<0
x^2+7x-8<0
(x+8)(x-1)<0
Critical values: x=-8 and x=1
Test intervals: x<-8 , -8<x<1 , x>1
Testing x=-10 for x<-8 :
(-10)^2+7(-10)-24=100-70-24=6
6 is not less than 0 , so x<-8 is not part of the solution set.
Testing x=0 for -8<x<1 :
(0)^2+7(0)-8=-8
-8 is less than 0 , so -8<x<1 is part of the solution set.
Testing x=2 for x>1 :
(2)^2+7(2)-8=4+14-8=10
10 is not less than 0 , so x>1 is not part of the solution set.
Testing the critical values, x=-8 and x=1 :
(-8)^2+7(-8)-8=0
(1)^2+7(1)-8=0
0 is not less than 0 , so neither -8 nor 1 is part of the solution set.
That means that the solution is -8<x<1

f) x^2-11x+9>-9

x^2-11x+9>-9
x^2-11x+18>0
(x-9)(x-2)>0
Critical values: x=9 and x=2
Test intervals: x<2 , 2<x<9 , 9<x
Testing x=0 for x<2 :
(0)^2-11(0)+18=18
18 is greater than 0 , so x<2 is part of the solution set.
Testing x=3 for 2<x<9 :
(3)^2-11(3)+18=9-33+18=-6
-6 is less than 0 , so 2<x<9 is not part of the solution set.
Testing x=10 for 9<x :
(10)^2-11(10)+18=100-110+18=8
8 is greater than 0 , so 9<x is part of the solution set.
That means that the solution is x<2\or9<x