Exercise 104 Page 153

Practice makes perfect

a In the given expression, the left-hand side is an absolute value. |5x+8|≥ -4 Absolute values are never negative. Therefore, |5x+8| is always greater than or equal to 0, which means the inequality holds for all real values of x.

b We are asked to solve the following inequality.

x^2+x-20<0 The solution to an inequality is a set of values. To find the boundary points of that set, we can replace the inequality sign with an equals sign. x^2+x-20=0Let's solve this quadratic equation using the Quadratic Formula.

x^2+x-20=0

UseQuadForm

Use the Quadratic Formula: a = 1, b= 1, c= -20

x=- 1±sqrt(1^2-4( 1)( -20))/2( 1)

▼

Evaluate right-hand side

BaseOne

1^a=1

x=- 1±sqrt(1-4(1)(-20))/2(1)

Multiply

x=- 1±sqrt(1+80)/2

AddTerms

Add terms

x=- 1±sqrt(81)/2

CalcRoot

Calculate root

x=-1±9/2

Let's split the solutions into positive and negative cases.

x=-1±9/2
x=-1-9/2	x=-1+9/2
x=-5	x=4

The boundaries for the inequality are x=-5 and x=4. Notice that the boundary points are open since the inequality is strict. This means the boundaries are not part of the solution set.

Let's choose a test value in each of the three regions and examine which of them solves the inequality.

x	x^2+x-20 < 0	Evaluate	True?
-6	( -6)^2+( -6)-20 ? < 0	10 ≮ 0	*
0	0^2+( 0)-20 ? < 0	-20 < 0	✓
5	5^2+( 5)-20 ? < 0	10 ≮ 0	*

As we can see, the region between the boundary points solves the inequality.

c We are given a quadratic equation.

2x^2-6x=-5 We can solve it using the Quadratic Formula. Let's start by rewriting it into standard form. Then, we can solve it.

2x^2-6x=-5

AddEqn

LHS+5=RHS+5

2x^2-6x+5=0

UseQuadForm

Use the Quadratic Formula: a = 2, b= -6, c= 5

x=- ( -6)±sqrt(( -6)^2-4( 2)( 5))/2( 2)

▼

Simplify right-hand side

NegNeg

- (- a)=a

x=6±sqrt((-6)^2-4(2)(5))/2(2)

NegBaseToPosBase

(- a)^2=a^2

x=6±sqrt(36-4(2)(5))/2(2)

Multiply

x=6±sqrt(36-40)/4

SubTerm

Subtract term

x=6±sqrt(-4)/4 *

Here we get the square root of -4 in the numerator. The square root of a negative number is not a real number. Therefore the equation has no real solutions.

d All terms in this equation are fractions.

5/9-x/3=4/9If we multiply both sides of the equation by 9, we can transform it to an easier equivalent equation.

5/9-x/3=4/9

MultEqn

LHS * 9=RHS* 9

9(5/9-x/3)=9* 4/9

▼

Simplify

Distr

Distribute 9

9*5/9-9*x/3=9*4/9

DenomMultFracToNumber