Core Connections Geometry, 2013
CC
Core Connections Geometry, 2013 View details
2. Section 12.2
Continue to next subchapter

Exercise 65 Page 753

Practice makes perfect
a
To solve an equation, we should first gather all of the variable terms on one side and all of the constant terms on the other side using the Properties of Equality. In this case, we need to start by using the Distributive Property to simplify the left-hand side of the equation. 4(x-3)=11 ⇕ 4x-12=11 Now, we can continue to solve using the Properties of Equality.
4x-12=11
4x=23
x = 23/4
b Let's take a closer look at the given equation.

x^2 = -10 Notice that if there is a real number, which is a solution to the equation above, its square must be a negative number. That however is not possible for real numbers and for this reason this equation has no real solution.

c To solve the given equation by taking the square roots, we need to consider the positive and negative solutions.
3 x^2 - 18 = 0
3 x^2 = 18
x^2 = 6
sqrt(x^2)=sqrt(6)

sqrt(a^2)=± a

x=± sqrt(6)
We found that x=± sqrt(6). Thus, there are two solutions for the equation, which are x=sqrt(6) and x=- sqrt(6).
d Let's take a closer look at the given equation.

-7 = | x-6 | In order to solve it, we would have to find some x such that the absolute value of x-6 would be negative. That however is impossible, since absolute value of any number is non-negative.