McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
5. Radical Equations
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Exercise 4 Page 261

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

4

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!
5+sqrt(g-3)=6
sqrt(g-3)=1
(sqrt(g-3))^2=1^2
g-3=1^2
g-3=1
g=4
The solution of our equation is g= 4. Now, let's check whether our solution is extraneous.

Checking the Solution

To check our solution, we will substitute 4 for g into the original equation. If we obtain a true statement, the solution is not extraneous. Otherwise, the solution is extraneous.
5+sqrt(g-3)=6
5+sqrt(4-3)? =6
â–Ľ
Simplify
5+sqrt(1)? =6
5+1? =6
6=6 âś“
We obtained a true statement, so g=4 is a solution to the equation.