Big Ideas Math Algebra 1, 2015
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Big Ideas Math Algebra 1, 2015 View details
3. Solving Radical Equations
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Exercise 79 Page 566

Start with the answer and use the Properties of Equality to build a new radical equation.

Example Solution: sqrt(x+11)=4

Practice makes perfect
We are asked to write a radical equation that has a solution of x= 5. To do so, we will start with the answer equation and then build a radical equation by using the Properties of Equality. x= 5One possible first step is to add or subtract an arbitrary number on both sides of the equation. We will choose to add 11 because when it is added to 5, we get a perfect square number 16 on the right-hand side. This will make it easier when we incorporate the radical in our equation. x+ 11= 5+ 11 ⇔ x+ 11= 16 Now, we will take the square root of both sides to make our obtained equation into a radical equation. sqrt(x+ 11)=sqrt(16) ⇔ sqrt(x+ 11)=4 Great! We created an equation that has a solution of x= 5. Note that we can write many different radical equations that have x= 5 as their solutions by using the Properties of Equality.

Checking Our Answer

Solving the Equation
Let's solve our equation to verify that its solution is x= 5.
sqrt(x+11)=4
â–Ľ
Solve for x
(sqrt(x+11))^2=4^2
x+11=4^2
x+11=16
x=5
Next, we will substitute it back into our equation to check that it is not an extraneous solution.
sqrt(x+11)=4
sqrt(5+11) ? = 4
sqrt(16) ? = 4
4 = 4 âś“
Because our substitution lead to an identity, we know that x= 5 is the real solution of our equation.