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

SqrtEqn

sqrt(LHS)=sqrt(RHS)

x=± sqrt(144)

CalcRoot

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

AddEqn

LHS+5=RHS+5

x^2=144

SqrtEqn

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)

AddEqn

LHS+1=RHS+1

x=1+sqrt(144)

CalcRoot

Calculate root

x=1+12

AddTerms

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

SubEqn

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.