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Core Connections Integrated II, 2015

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Core Connections Integrated II, 2015 View details

2. Section 5.2

Finish the assignment

Continue to next subchapter

Exercise 113 Page 301

Practice makes perfect

a

To solve the equation, we have to perform inverse operations until x is isolated. Note that taking an even indexed root out of an equation gives two possible solutions — a positive and a negative one.

x^2=144

sqrt(LHS)=sqrt(RHS)

x=± sqrt(144)

Calculate root

x=± 12

b

Examining the equation we notice that x is already isolated, which means the equations is already solved. However, we can calculate the right-hand side by simplifying the radical.

x=sqrt(144) ⇔ x=12

c

Like in Part A, we will perform inverse operations until x is isolated.

x^2-5=139

LHS+5=RHS+5

x^2=144

sqrt(LHS)=sqrt(RHS)

x=± 12

d

Like in previous parts, we will solve for x by performing inverse operations until x is isolated.

x-1=sqrt(144)

LHS+1=RHS+1

x=1+sqrt(144)

Calculate root

x=1+12

Add terms

x=13

e

The square of a number is always non-negative. This means that the given equation cannot be true, since it tells us that the square of x equals -144.

x^2≠ -144 Therefore, the equation has no solutions.

f

Let's start by isolating x^2 on the left-hand side.

x^2+5=0

LHS-5=RHS-5

x^2=-5

Like in Part E, this is also an impossible equation. x^2≠ -5 Therefore, the equation has no solutions.

Subchapter links

5.2.1 Perfect Square Equations p.285-286

5.2.2 Completing the Square p.289

5.2.3 More Completing the Square p.291-292

5.2.4 Introduction to the Quadratic Formula p.296-297

5.2.5 Solving and Applying Quadratic Equations p.301-304

5.2.6 Introducing Complex Numbers p.308-310