Exercise 29 Page &nbsp; 740

u= -1

2/3(15( -1)-6)? =14( -1)

MultPosNeg

a(- b)=- a * b

2/3(-15-6)? =-14

SubTerm

Subtract term

2/3(-21)? =-14

MultPosNeg

a(- b)=- a * b

-14 = -14 ✓

Since the substitution resulted in a true equality, our answer is correct.

b We want to find the roots of the given equation.

(5-x)(3x+8)=0 Since the equation is already written in factored form, we will use the Zero Product Property.

(5-x)(3x+8)=0

ZeroProdProp

Use the Zero Product Property

lc5-x=0 & (I) 3x+8=0 & (II)

AddEqn

(I): LHS+x=RHS+x

l5=x 3x+8=0

RearrangeEqn

(I): Rearrange equation

lx=5 3x+8=0

SubEqn

(II): LHS-8=RHS-8

lx=5 3x=-8

DivEqn

(II): .LHS /3.=.RHS /3.

lx=5 x= -83

MoveNegNumToFrac

(II): Put minus sign in front of fraction

lx=5 x=- 83

We found that the roots of the equation are x = 5 and x = - 83. In order to check whether our answers are correct, we need to substitute them for x in the original equation. Let's start by checking x = 5.

(5-x)(3x+8)=0

x= 5

(5- 5)(3( 5)+8)? =0

(5-5)(15+8)? =0

AddSubTerms

Add and subtract terms

0(23)? =0

ZeroPropMult

Zero Property of Multiplication

0=0 ✓

Since the substitution resulted in a true equality, the solution is correct. Next, let's substitute - 83 for x in the original equation.

(5-x)(3x+8)=0

x= -8/3

(5-( -8/3))(3( -8/3)+8)? =0

SubNeg

a-(- b)=a+b

(5+8/3)(3(-8/3)+8)? =0

MoveNegFracToNum

Put minus sign in numerator

(5+8/3)(3(-8/3)+8)? =0

DenomMultFracToNumber

3 * a/3= a

(5+8/3)(-8+8)? =0

WriteAsFrac

Write as a fraction

(15/3+8/3)(-8+8)? =0

AddTerms

Add terms

23/3(0)? =0

ZeroPropMult

Zero Property of Multiplication

0=0 ✓

Again, since the substitution resulted in a true equality, the solution is correct.

c We want to find the roots of the given equation.

2(k-5)^2=32 In order to do so, we will first use change it into an equivalent absolute value equation.

2(k-5)^2=32

DivEqn

.LHS /2.=.RHS /2.

(k-5)^2 = 16

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt((k-5)^2) = sqrt(16)

CalcRoot

Calculate root

sqrt((k-5)^2) = 4

SqrtToAbs

sqrt(a^2)=|a|

| k-5 | = 4

An absolute value measures an expression's distance from a midpoint on a number line. |k-5|= 4 This equation means that the distance is 4, either in the positive direction or the negative direction. |k-5|= 4 ⇒ lk-5= 4 k-5= - 4 To find the solutions to the absolute value equation, we need to solve both of these cases for k.

| k-5|=4

lc k-5 ≥ 0:k-5 = 4 & (I) k-5 < 0:k-5 = - 4 & (II)

lck-5=4 & (I) k-5=- 4 & (II)

(I), (II): LHS+5=RHS+5

lk_1=9 k_2=1

Both 9 and 1 are solutions to the absolute value equation as well as the original equation. In order to check whether our answers are correct, we need to substitute them for k in the original equation. Let's start by checking x = 1.

2(k-5)^2=32

x= 1

2( 1-5)^2? =32

SubTerm

Subtract term

2(-4)^2? =32

Calculate power

2(16)? =32

32=32 ✓

Since the substitution resulted in a true equality, the solution is correct. Next, let's substitute 9 for k in the original equation.

2(k-5)^2=32

x= 1

2( 9-5)^2? =323)+8)? =0

SubTerm

Subtract term

2(4)^2? =32

Calculate power

2(16)? =32

32=32 ✓

Again, since the substitution resulted in a true equality, the solution is correct.

We will use the Quadratic Formula to solve the given quadratic equation. ax^2+ bx+ c=0 ⇕ x=- b± sqrt(b^2-4 a c)/2 a We first need to identify the values of a, b, and c. 2p^2+7p-9=0 ⇕ 2p^2+ 7p+( - 9)=0 We see that a= 2, b= 7, and c= - 9. Let's substitute these values into the Quadratic Formula.

p=- b±sqrt(b^2-4ac)/2a

SubstituteValues

Substitute values

p=- 7±sqrt(7^2-4( 2)( - 9))/2( 2)

▼

Solve for p and Simplify

Calculate power

p=-7±sqrt(49-4(2)(- 9))/2(2)

p=-7±sqrt(49-8(- 9))/4

MultNegNegOnePar

- a(- b)=a* b

p=-7±sqrt(49+72)/4

AddTerms

Add terms

p=-7±sqrt(121)/4

CalcRoot

Calculate root

p=-7 ± 11/4

The solutions for this equation are p= -7 ± 114. Let's separate them into the positive and negative cases.

p=-7 ± 11/4
p_1=-7 + 11/4	p_2=-7 - 11/4
p_1=4/4	p_2=-18/4
p_1=1	p_2=-4.5

Using the Quadratic Formula, we found that the solutions of the given equation are p_1=1 and p_2=-4.5. In order to check whether our answers are correct, we need to substitute them for p in the original equation. Let's start by checking p = 1.

2p^2 + 7p-9 = 0

p= 1

2( 1)^2 + 7( 1)-9? =0