McGraw Hill Integrated II, 2012
MH
McGraw Hill Integrated II, 2012 View details
5. Radical Equations
Continue to next subchapter

Exercise 3 Page 261

Rearrange the radical equation so that the radical expression is isolated. Then raise both sides of the equation to the second power.

2

Practice makes perfect

We will find and check the solution of the given equation.

Finding the Solution

To solve an equation with a variable expression inside a radical, we will first rearrange the radical equation so that the radical expression is isolated. Then, we can raise both sides of the equation to a power equal to the index of the radical. In this case, we will raise both sides of the equation to the second power. Let's do it!
sqrt(7r+2)+3=7
sqrt(7r+2)=4
(sqrt(7r+2))^2=4^2
7r+2=4^2
7r+2=16
7r=14
r=2
The solution of our equation is r= 2. Now, let's check whether our solution is extraneous.

Checking the Solution

To check our solution, we will substitute 2 for r into the original equation. If we obtain a true statement, the solution is not extraneous. Otherwise, the solution is extraneous.
sqrt(7r+2)+3=7
sqrt(7( 2)+2)+3? =7
â–Ľ
Simplify
sqrt(14+2)+3? =7
sqrt(16)+3? =7
4+3? =7
7=7 âś“
We obtained a true statement, so r=2 is a solution to the equation.