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 4 Page 559

Practice makes perfect
a We are asked to solve the following square root equation.
5 = sqrt(x+20) To do so, let's recall the steps that we listed in the previous exercise.
  1. Isolate the radical on one side of the equation.
  2. Square both sides of the equation.
  3. Solve the equation for the variable.
  4. Substitute the solution into the original equation to check whether it is extraneous or not.
Since our radical is already isolated, we will continue by squaring both sides of the equation. Then, we will solve the equation for the variable x.
5= sqrt(x+20)
5^2=( sqrt(x+20))^2
25= x+20
5= x
x=5
Great! Let's finally check the result by substituting x= 5 back into the original equation!
5=sqrt(x+20)
5 ? = sqrt(5+20)
5 ? = sqrt(25)
5 = 5 âś“
Since we found an identity, we know that the solution to the equation is x=5.
b We will solve the next square root equation by following the same steps that we took in Part A.
4 = sqrt(x-18)Having an isolated radical, we can immediately square both sides of the equation and then solve it for x.
4= sqrt(x-18)
4^2=( sqrt(x-18))^2
16= x-18
34= x
x=34
At last, we will substitute the solution back into the original equation to check the extraneous solutions.
4=sqrt(x-18)
4 ? = sqrt(34-18)
4 ? = sqrt(16)
4 = 4 âś“
Great! We ended with an identity again. Therefore, the result is x=34.
c Let's take a look at the given equation.
sqrt(x)+2 = 3One more time, we will solve it by applying the same steps as listed in Part A. This time, we will first isolate the radical on one side and then square both sides of the equation. Let's do it!
sqrt(x)+2=3
sqrt(x)=1
( sqrt(x))^2=1^2
x=1
Now, we will check the result.
sqrt(x)+2=3
sqrt(1)+2 ? = 3
1+2 ? = 3
3 = 3 âś“
Having an identity, we can be sure that the solution is x=1.
d One last time, we will solve the given square root equation.
-3 = - 2 sqrt(x)Let's proceed in the same way as before. We will first isolate the radical and then square both sides of the equation.
-3= -2 sqrt(x)
-3/-2 = sqrt(x)
3/2 = sqrt(x)
(3/2 )^2 = ( sqrt(x) )^2
9/4= x
x=9/4
Next, let's check the solution by substituting x= 94 back into the original equation!
-3=-2 sqrt(x)
-3 ? = -2 sqrt(9/4)
â–Ľ
Evaluate right-hand side
-3 ? = -2 * 3/2
-3 ? = -6/2
- 3 = - 3 âś“
Great! Finally, we can say that x= 94 satisfies the equation.