Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
1. Section 10.1
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Exercise 32 Page 558

Practice makes perfect
a Before we can solve this equation, we have to distribute the factor outside of the parentheses on the left-hand side. After that, we can perform inverse operations until x is isolated.
4(2x+5)-11=4x-3
8x+20-11=4x-3
8x+9=4x-3
8x=4x-12
4x=- 12
x=- 3
b By multiplying both sides of the equation by both denominators, we can eliminate the fractions and then proceed to solve the equation.
2m-1/19=m/10
2m-1=19* m/10
10(2m-1)=19m
20m-10=19m
m-10=0
m=10
c To solve this equation we will use the Quadratic Formula.
p=- b± sqrt(b^2-4ac)/2a

a= 3, b= 10, c= - 8

p=- 10± sqrt(10^2-4( 3)( - 8))/2( 3)
p=- 10± sqrt(100-4(3)(- 8))/2(3)
p=- 10± sqrt(100+96)/6
p=- 10± sqrt(196)/6
p=- 10± 14/6
Finally, to find the roots we have to split the fraction into the positive and negative case. Positive:& - 10 + 14/6= 2/3 [0.5em] Negative:& - 10 - 14/6=- 4 The equations roots are p=- 4 and p= 23.
d Since the inverse operation to a square root is to square, we have to square both sides to eliminate the square root.
sqrt(x+2)=5
(sqrt(x+2))^2=5^2
x+2=5^2
x+2=25
x=23
Note that squaring an equation can possibly introduce false roots. Therefore, we must test our solution by substituting it into the original equation and then simplify.
sqrt(x+2)=5
sqrt(23+2)? =5
sqrt(25)? =5
5=5
The solution is correct.